686. Oxidation-Reduction Reactions -
Earth's atmosphere contains about $20 \%$ molecular oxygen, $\mathrm{O}_{2}$, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving $\mathrm{O}_{2}$, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification.

Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:

$$
2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)
$$

It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:

$$
\begin{aligned}
& 2 \mathrm{Na}(s) \longrightarrow 2 \mathrm{Na}^{+}(s)+2 \mathrm{e}^{-} \\
& \mathrm{Cl}_{2}(g)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}(s)
\end{aligned}
$$

These equations show that Na atoms lose electrons while Cl atoms (in the $\mathrm{Cl}_{2}$ molecule) gain electrons, the " s " subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:

```
oxidation = loss of electrons
reduction = gain of electrons
```

In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium.


686. Oxidation-Reduction Reactions -
$$
\begin{aligned}
\text { reducing agent } & =\text { species that is oxidized } \\
\text { oxidizing agent } & =\text { species that is reduced }
\end{aligned}
$$

Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl :

$$
\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g)
$$

The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound were ionic. The following guidelines are used to assign oxidation numbers to each element in a molecule or ion.

1. The oxidation number of an atom in an elemental substance is zero.
2. The oxidation number of a monatomic ion is equal to the ion's charge.
3. Oxidation numbers for common nonmetals are usually assigned as follows:

- Hydrogen: +1 when combined with nonmetals, -1 when combined with metals
- Oxygen: - 2 in most compounds, sometimes -1 (so-called peroxides, $\mathrm{O}_{2}{ }^{2-}$ ), very rarely $-\frac{1}{2}$ (so-called superoxides, $\mathrm{O}_{2}{ }^{-}$), positive values when combined with F (values vary)
- Halogens: -1 for $F$ always, -1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)

4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., $2+$ ), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2 ). This convention aims to emphasize the distinction between these two related properties.


688. Assigning Oxidation Numbers -
Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:
(a) $\mathrm{H}_{2} \mathrm{~S}$
(b) $\mathrm{SO}_{3}{ }^{2-}$
(c) $\mathrm{Na}_{2} \mathrm{SO}_{4}$


689. Solution -
(a) According to guideline 3 , the oxidation number for H is +1 .

Using this oxidation number and the compound's formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

$$
\begin{gathered}
\text { charge on } \mathrm{H}_{2} \mathrm{~S}=0=(2 \times+1)+(1 \times x) \\
x=0-(2 \times+1)=-2
\end{gathered}
$$

(b) Guideline 3 suggests the oxidation number for oxygen is -2 .

Using this oxidation number and the ion's formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

$$
\begin{gathered}
\text { charge on } \mathrm{SO}_{3}{ }^{2-}=-2=(3 \times-2)+(1 \times x) \\
x=-2-(3 \times-2)=+4
\end{gathered}
$$

(c) For ionic compounds, it's convenient to assign oxidation numbers for the cation and anion separately.

According to guideline 2, the oxidation number for sodium is +1 .
Assuming the usual oxidation number for oxygen ( -2 per guideline 3 ), the oxidation number for sulfur is calculated as directed by guideline 4:

$$
\begin{gathered}
\text { charge on } \mathrm{SO}_{4}{ }^{2-}=-2=(4 \times-2)+(1 \times x) \\
x=-2-(4 \times-2)=+6
\end{gathered}
$$


690. Check Your Learning -
Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:
(a) $\mathrm{KNO}_{3}$
(b) $\mathrm{AlH}_{3}$
(c) $\mathrm{NH}_{4}{ }^{+}$
(d) $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$

Answer:
(a) N, +5; (b) Al, +3; (c) N, -3; (d) P, +5

Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. (While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist Example 7.6.) Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:

$$
\begin{aligned}
\text { oxidation } & =\text { increase in oxidation number } \\
\text { reduction } & =\text { decrease in oxidation number }
\end{aligned}
$$

Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl ) and chlorine is reduced (its oxidation number decreases from 0 in $\mathrm{Cl}_{2}$ to -1 in NaCl ). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in $\mathrm{H}_{2}$ to +1 in HCl ) and chlorine is reduced (its oxidation number decreases from 0 in
$\mathrm{Cl}_{2}$ to -1 in HCl ).
Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions such as the one depicted in Figure 7.1 are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:


690. Check Your Learning -
$$
10 \mathrm{Al}(s)+6 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) \longrightarrow 4 \mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{AlCl}_{3}(s)+12 \mathrm{H}_{2} \mathrm{O}(g)+3 \mathrm{~N}_{2}(g)
$$


691. LINK TO LEARNING -
Watch a brief video (http://openstax.org/1/16hybridrocket) showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at
3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.

Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:

$$
\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)
$$

Metallic elements may also be oxidized by solutions of other metal salts; for example:

$$
\mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)
$$

This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting $\mathrm{Cu}^{2+}$ ions dissolve in the solution to yield a characteristic blue color (Figure 7.9).

(a)

(b)

(c)

FIGURE 7.9 (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire and development of a blue color in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)


693. Describing Redox Reactions -
Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.
(a) $\mathrm{ZnCO}_{3}(s) \longrightarrow \mathrm{ZnO}(s)+\mathrm{CO}_{2}(g)$
(b) $2 \mathrm{Ga}(l)+3 \mathrm{Br}_{2}(l) \longrightarrow 2 \mathrm{GaBr}_{3}(s)$
(c) $2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)$
(d) $\mathrm{BaCl}_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{BaSO}_{4}(s)+2 \mathrm{KCl}(a q)$
(e) $\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$


694. Solution -
Redox reactions are identified per definition if one or more elements undergo a change in oxidation number.
(a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
(b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in $\mathrm{Ga}(I)$ to +3 in $\operatorname{GaBr}_{3}(S)$. The reducing agent is $\mathrm{Ga}(I)$. Bromine is reduced, its oxidation number decreasing from 0 in $\mathrm{Br}_{2}(I)$ to -1 in $\operatorname{GaBr}_{3}(S)$. The oxidizing agent is $\mathrm{Br}_{2}(I)$.
(c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from -1 in $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})$ to 0 in $\mathrm{O}_{2}(\mathrm{~g})$. Oxygen is also reduced, its oxidation number decreasing from -1 in $\mathrm{H}_{2} \mathrm{O}_{2}(a q)$ to -2 in $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$. For disproportionation reactions, the same substance functions as an oxidant and a reductant.
(d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
(e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from -2 in $\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})$ to +4 in $\mathrm{CO}_{2}(\mathrm{~g})$. The reducing agent (fuel) is $\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})$. Oxygen is reduced, its oxidation number decreasing from 0 in $\mathrm{O}_{2}(\mathrm{~g})$ to -2 in $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$. The oxidizing agent is $\mathrm{O}_{2}(\mathrm{~g})$.


695. Check Your Learning -
This equation describes the production of tin(II) chloride:

$$
\mathrm{Sn}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{SnCl}_{2}(s)+\mathrm{H}_{2}(g)
$$

Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.


696. Answer: -
Yes, a single-replacement reaction. $\mathrm{Sn}(s)$ is the reductant, $\mathrm{HCl}(g)$ is the oxidant.


697. Balancing Redox Reactions via the Half-Reaction Method -
Redox reactions that take place in aqueous media often involve water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the method of half-reactions, which involves the following steps:

1. Write the two half-reactions representing the redox process.
2. Balance all elements except oxygen and hydrogen.
3. Balance oxygen atoms by adding $\mathrm{H}_{2} \mathrm{O}$ molecules.
4. Balance hydrogen atoms by adding $\mathrm{H}^{+}$ions.
5. Balance charge by adding electrons.
6. If necessary, multiply each half-reaction's coefficients by the smallest possible integers to yield equal numbers of electrons in each.
7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.
8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:
a. Add $\mathrm{OH}^{-}$ions to both sides of the equation in numbers equal to the number of $\mathrm{H}^{+}$ions.
b. On the side of the equation containing both $\mathrm{H}^{+}$and $\mathrm{OH}^{-}$ions, combine these ions to yield water molecules.
c. Simplify the equation by removing any redundant water molecules.
9. Finally, check to see that both the number of atoms and the total charges ${ }^{\frac{1}{}}$ are balanced.


699. Balancing Redox Reactions in Acidic Solution -
Write a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution.

$$
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+}
$$


700. Solution -
Step 1.
Write the two half-reactions.
Each half-reaction will contain one reactant and one product with one element in common.

$$
\begin{gathered}
\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} \\
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+}
\end{gathered}
$$

Step 2.
Balance all elements except oxygen and hydrogen. The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms.

$$
\begin{gathered}
\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} \\
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}
\end{gathered}
$$

Step 3.
Balance oxygen atoms by adding $\mathrm{H}_{2} \mathrm{O}$ molecules. The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side.

$$
\begin{gathered}
\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} \\
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
\end{gathered}
$$

Step 4.
Balance hydrogen atoms by adding $\mathrm{H}^{+}$ions. The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side.


700. Solution -
$$
\begin{gathered}
\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} \\
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
\end{gathered}
$$

Step 5.
Balance charge by adding electrons. The iron half-reaction shows a total charge of $2+$ on the left side (1 $\mathrm{Fe}^{2+}$ ion) and $3+$ on the right side ( $1 \mathrm{Fe}^{3+}$ ion). Adding one electron to the right side brings that side's total charge to $(3+)+(1-)=2+$, and charge balance is achieved.

The chromium half-reaction shows a total charge of $(1 \times 2-)+(14 \times 1+)=12+$ on the left side $\left(1 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right.$ ion and $14 \mathrm{H}^{+}$ions). The total charge on the right side is $(2 \times 3+)=6+\left(2 \mathrm{Cr}^{3+}\right.$ ions $)$. Adding six electrons to
the left side will bring that side's total charge to $(12++6-)=6+$, and charge balance is achieved.

$$
\begin{gathered}
\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} \\
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
\end{gathered}
$$

Step 6.
Multiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction. To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction's coefficient must be multiplied by 6 .


700. Solution -
$$
\begin{gathered}
6 \mathrm{Fe}^{2+} \longrightarrow 6 \mathrm{Fe}^{3+}+6 \mathrm{e}^{-} \\
\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+6 \mathrm{e}^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
\end{gathered}
$$

Step 7.
Add the balanced half-reactions and cancel species that appear on both sides of the equation.

$$
6 \mathrm{Fe}^{2+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+6 \mathrm{e}^{-}+14 \mathrm{H}^{+} \longrightarrow 6 \mathrm{Fe}^{3+}+6 \mathrm{e}^{-}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
$$

Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:

$$
6 \mathrm{Fe}^{2+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+} \longrightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}
$$

A final check of atom and charge balance confirms the equation is balanced.

|  | Reactants | Products |
| :--- | :--- | :--- |
| Fe | 6 | 6 |
| Cr | 2 | 2 |
| O | 7 | 7 |
| H | 14 | 14 |
| charge | $24+$ | $24+$ |


701. Check Your Learning -
In basic solution, molecular chlorine, $\mathrm{Cl}_{2}$, reacts with hydroxide ions, $\mathrm{OH}^{-}$, to yield chloride ions, $\mathrm{Cl}^{-}$. and chlorate ions, $\mathrm{ClO}_{3}{ }^{-}$. HINT: This is a disproportionation reaction in which the element chlorine is both oxidized and reduced. Write a balanced equation for this reaction.


702. Answer: -
$3 \mathrm{Cl}_{2}(a q)+6 \mathrm{OH}^{-}(a q) \longrightarrow 5 \mathrm{Cl}^{-}(a q)+\mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)$


703. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Explain the concept of stoichiometry as it pertains to chemical reactions
- Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products
- Perform stoichiometric calculations involving mass, moles, and solution molarity

A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing
classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction's stoichiometry, a term derived from the Greek words stoicheion (meaning "element") and metron (meaning "measure"). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.

The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, $\frac{3}{4}$ cup milk, and one egg. The "equation" representing the preparation of pancakes per this recipe is

$$
1 \text { cup mix }+\frac{3}{4} \text { cup milk }+1 \text { egg } \longrightarrow 8 \text { pancakes }
$$

If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is

$$
24 \text { pancakes } \times \frac{1 \mathrm{egg}}{8 \text { pancakes- }}=3 \mathrm{eggs}
$$

Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:


703. LEARNING OBJECTIVES -
$$
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)
$$

This equation shows ammonia molecules are produced from hydrogen molecules in a $2: 3$ ratio, and stoichiometric factors may be derived using any amount (number) unit:

$$
\frac{2 \mathrm{NH}_{3} \text { molecules }}{3 \mathrm{H}_{2} \text { molecules }} \text { or } \frac{2 \text { doz } \mathrm{NH}_{3} \text { molecules }}{3 \text { doz } \mathrm{H}_{2} \text { molecules }} \text { or } \frac{2 \mathrm{~mol} \mathrm{NH}_{3} \text { molecules }}{3 \mathrm{~mol} \mathrm{H}_{2} \text { molecules }}
$$

These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.


705. Moles of Reactant Required in a Reaction -
How many moles of $\mathrm{I}_{2}$ are required to react with 0.429 mol of Al according to the following equation (see Figure 7.10)?

$$
2 \mathrm{Al}+3 \mathrm{I}_{2} \longrightarrow 2 \mathrm{AlI}_{3}
$$

FIGURE 7.10 Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)


706. Solution -
Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest
is $\frac{3 \mathrm{~mol} \mathrm{I}_{2}}{2 \mathrm{~mol} \mathrm{Al}}$. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:

$$
\begin{aligned}
\mathrm{mol} \mathrm{I}_{2} & =0.429 \mathrm{~mol} \mathrm{At} \times \frac{3 \mathrm{~mol} \mathrm{I}_{2}}{2 \mathrm{~mol} \mathrm{At}} \\
& =0.644 \mathrm{~mol} \mathrm{I}_{2}
\end{aligned}
$$


707. Check Your Learning -
How many moles of $\mathrm{Ca}(\mathrm{OH})_{2}$ are required to react with 1.36 mol of $\mathrm{H}_{3} \mathrm{PO}_{4}$ to produce $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ according to the equation $3 \mathrm{Ca}(\mathrm{OH})_{2}+2 \mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+6 \mathrm{H}_{2} \mathrm{O}$ ?


708. Answer: -
2.04 mol


710. Number of Product Molecules Generated by a Reaction -
How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?

$$
\mathrm{C}_{3} \mathrm{H}_{8}+5 \mathrm{O}_{2} \longrightarrow 3 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}
$$


711. Solution -
The approach here is the same as for Example 7.8, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro's number.

The balanced equation shows that carbon dioxide is produced from propane in a $3: 1$ ratio:

$$
\frac{3 \mathrm{~mol} \mathrm{CO}_{2}}{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}}
$$

Using this stoichiometric factor, the provided molar amount of propane, and Avogadro's number,


712. Check Your Learning -
How many $\mathrm{NH}_{3}$ molecules are produced by the reaction of 4.0 mol of $\mathrm{Ca}(\mathrm{OH})_{2}$ according to the following equation:

$$
\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}+\mathrm{Ca}(\mathrm{OH})_{2} \longrightarrow 2 \mathrm{NH}_{3}+\mathrm{CaSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}
$$


713. Answer: -
$4.8 \times 10^{24} \mathrm{NH}_{3}$ molecules

These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an
easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.


715. Relating Masses of Reactants and Products -
What mass of sodium hydroxide, NaOH , would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, $\mathrm{Mg}(\mathrm{OH})_{2}$ ] by the following reaction?

$$
\mathrm{MgCl}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{NaCl}(a q)
$$


716. Solution -
The approach used previously in Example 7.8 and Example 7.9 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:


717. Check Your Learning -
What mass of gallium oxide, $\mathrm{Ga}_{2} \mathrm{O}_{3}$, can be prepared from 29.0 g of gallium metal? The equation for the reaction is $4 \mathrm{Ga}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{Ga}_{2} \mathrm{O}_{3}$.


718. Answer: -
39.0 g


720. Relating Masses of Reactants -
What mass of oxygen gas, $\mathrm{O}_{2}$, from the air is consumed in the combustion of 702 g of octane, $\mathrm{C}_{8} \mathrm{H}_{18}$, one of the principal components of gasoline?

$$
2 \mathrm{C}_{8} \mathrm{H}_{18}+25 \mathrm{O}_{2} \longrightarrow 16 \mathrm{CO}_{2}+18 \mathrm{H}_{2} \mathrm{O}
$$


721. Solution -
The approach required here is the same as for the Example 7.10, differing only in that the provided and requested masses are both for reactant species.


722. Check Your Learning -
What mass of CO is required to react with 25.13 g of $\mathrm{Fe}_{2} \mathrm{O}_{3}$ according to the equation
$\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \longrightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2}$ ?


723. Answer: -
13.22 g

These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure 7.11 provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.

FIGURE 7.11 The flowchart depicts the various computational steps involved in most reaction stoichiometry
calculations.


725. Airbags -
Airbags (Figure 7.12) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, $\mathrm{NaN}_{3}$. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of $\mathrm{NaN}_{3}$ to initiate its decomposition:

$$
2 \mathrm{NaN}_{3}(s) \longrightarrow 3 \mathrm{~N}_{2}(g)+2 \mathrm{Na}(s)
$$

This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second ( $\sim 0.03-0.1 \mathrm{~s}$ ). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass $(\sim 100 \mathrm{~g})$ of $\mathrm{NaN}_{3}$ will generate approximately 50 L of $\mathrm{N}_{2}$.

FIGURE 7.12 Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)


726. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Explain the concepts of theoretical yield and limiting reactants/reagents.
- Derive the theoretical yield for a reaction under specified conditions.
- Calculate the percent yield for a reaction.

The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.


727. Limiting Reactant -
Consider another food analogy, making grilled cheese sandwiches (Figure 7.13):
1 slice of cheese +2 slices of bread $\longrightarrow 1$ sandwich
Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio.

Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.

FIGURE 7.13 Sandwich making can illustrate the concepts of limiting and excess reactants.
Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:

$$
\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g)
$$

The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of $\mathrm{H}_{2}$ and 2 moles of $\mathrm{Cl}_{2}$. This represents a $3: 2$ (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine ( 2 mol ) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted.


727. Limiting Reactant -
An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction's stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield

$$
\text { mol } \mathrm{HCl} \text { produced }=3 \mathrm{~mol} \mathrm{H}_{2} \times \frac{2 \mathrm{~mol} \mathrm{HCl}}{1 \mathrm{~mol} \mathrm{H}_{2}}=6 \mathrm{~mol} \mathrm{HCl}
$$

Complete reaction of the provided chlorine would produce

$$
\text { mol } \mathrm{HCl} \text { produced }=2 \mathrm{~mol} \mathrm{Cl}_{2} \times \frac{2 \mathrm{~mol} \mathrm{HCl}}{1 \mathrm{~mol} \mathrm{Cl}_{2}}=4 \mathrm{~mol} \mathrm{HCl}
$$

The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl , there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure 7.14).

FIGURE 7.14 When $\mathrm{H}_{2}$ and $\mathrm{Cl}_{2}$ are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant.


728. LINK TO LEARNING -
View this interactive simulation (http://openstax.org/l/16reactantprod) illustrating the concepts of limiting and excess reactants.


730. Identifying the Limiting Reactant -
Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:

$$
3 \mathrm{Si}(s)+2 \mathrm{~N}_{2}(g) \longrightarrow \mathrm{Si}_{3} \mathrm{~N}_{4}(s)
$$

Which is the limiting reactant when 2.00 g of Si and 1.50 g of $\mathrm{N}_{2}$ react?


731. Solution -
Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.

$$
\begin{gathered}
\mathrm{mol} \mathrm{Si}=2.00 \mathrm{~g} \mathrm{Si} \times \frac{1 \mathrm{~mol} \mathrm{Si}}{28.09 \mathrm{~g} \mathrm{Si}}=0.0712 \mathrm{~mol} \mathrm{Si} \\
\mathrm{~mol} \mathrm{~N}_{2}=1.50-\mathrm{g}_{\mathrm{g}}^{\mathrm{g} \mathrm{~N}_{2}} \times \frac{1 \mathrm{~mol} \mathrm{~N}_{2}}{28.02-\frac{\mathrm{g} \mathrm{~N}_{2}}{}}=0.0535 \mathrm{~mol} \mathrm{~N}_{2}
\end{gathered}
$$

The provided $\mathrm{Si}: \mathrm{N}_{2}$ molar ratio is:

$$
\frac{0.0712 \mathrm{~mol} \mathrm{Si}}{0.0535 \mathrm{~mol} \mathrm{~N}_{2}}=\frac{1.33 \mathrm{~mol} \mathrm{Si}}{1 \mathrm{~mol} \mathrm{~N}_{2}}
$$

The stoichiometric Si: $\mathrm{N}_{2}$ ratio is:

$$
\frac{3 \mathrm{~mol} \mathrm{Si}}{2 \mathrm{~mol} \mathrm{~N}_{2}}=\frac{1.5 \mathrm{~mol} \mathrm{Si}}{1 \mathrm{~mol} \mathrm{~N}_{2}}
$$

Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant.

Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield


731. Solution -
Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield

$$
\mathrm{mol} \mathrm{Si}_{3} \mathrm{~N}_{4} \text { produced }=0.0712 \mathrm{~mol} \mathrm{Si} \times \frac{1 \mathrm{~mol} \mathrm{Si}_{3} \mathrm{~N}_{4}}{3 \mathrm{~mol} \mathrm{Si}}=0.0237 \mathrm{~mol} \mathrm{Si}_{3} \mathrm{~N}_{4}
$$

while the 0.0535 moles of nitrogen would produce

$$
\mathrm{mol} \mathrm{Si}_{3} \mathrm{~N}_{4} \text { produced }=0.0535 \mathrm{~mol} \mathrm{~N}_{2} \times \frac{1 \mathrm{~mol} \mathrm{Si}_{3} \mathrm{~N}_{4}}{2 \mathrm{~mol} \mathrm{~N}_{2}}=0.0268 \mathrm{~mol} \mathrm{Si}_{3} \mathrm{~N}_{4}
$$

Since silicon yields the lesser amount of product, it is the limiting reactant.


732. Check Your Learning -
Which is the limiting reactant when 5.00 g of $\mathrm{H}_{2}$ and 10.0 g of $\mathrm{O}_{2}$ react and form water?


733. Answer: -
$\mathrm{O}_{2}$


734. Percent Yield -
The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction's theoretical yield is achieved is commonly expressed as its percent yield:

$$
\text { percent yield }=\frac{\text { actual yield }}{\text { theoretical yield }} \times 100 \%
$$

Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.


736. Calculation of Percent Yield -
Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:

$$
\mathrm{CuSO}_{4}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{ZnSO}_{4}(a q)
$$

What is the percent yield?


737. Solution -
The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:

$$
1.274-\mathrm{g} \mathrm{CuSO}_{4} \times \frac{1 \mathrm{molCuSO}_{4}}{159.62-\mathrm{gCuSO}_{4}} \times \frac{1 \mathrm{molCu}}{1 \mathrm{molCuSO}_{4}} \times \frac{63.55 \mathrm{~g} \mathrm{Cu}}{1 \mathrm{~mol} \mathrm{Cu}}=0.5072 \mathrm{~g} \mathrm{Cu}
$$

Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be

$$
\begin{aligned}
\text { percent yield } & =\left(\frac{\text { actual yield }}{\text { theoretical yield }}\right) \times 100 \\
\text { percent yield } & =\left(\frac{0.392 \mathrm{~g} \mathrm{Cu}}{0.5072 \mathrm{~g} \mathrm{Cu}}\right) \times 100 \\
& =77.3 \%
\end{aligned}
$$


738. Check Your Learning -
excess HF?

$$
\mathrm{CCl}_{4}+2 \mathrm{HF} \longrightarrow \mathrm{CF}_{2} \mathrm{Cl}_{2}+2 \mathrm{HCl}
$$


739. Answer: -
48.3\%


741. Green Chemistry and Atom Economy -
The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the "Twelve Principles of Green Chemistry" (see details at this website (http://openstax.org/l/16greenchem)). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used:

$$
\text { atom economy }=\frac{\text { mass of product }}{\text { mass of reactants }} \times 100 \%
$$

Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.

The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure 7.15). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole ( 206 g ) of ibuprofen, an atom economy of $40 \%$. In the 1990 s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of $\sim 80 \%$, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency's Greener Synthetic Pathways Award in 1997.

(a)

(b)


741. Green Chemistry and Atom Economy -
(a)

(b)

FIGURE 7.15 (a) Ibuprofen is a popular nonprescription pain medication commonly sold as 200 mg tablets. (b) The BHC process for synthesizing ibuprofen requires only three steps and exhibits an impressive atom economy. (credit a: modification of work by Derrick Coetzee)


742. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the fundamental aspects of titrations and gravimetric analysis.
- Perform stoichiometric calculations using typical titration and gravimetric data.

In the 18th century, the strength (actually the concentration) of vinegar samples was determined by noting the amount of potassium carbonate, $\mathrm{K}_{2} \mathrm{CO}_{3}$, which had to be added, a little at a time, before bubbling ceased. The greater the weight of potassium carbonate added to reach the point where the bubbling ended, the more concentrated the vinegar.

We now know that the effervescence that occurred during this process was due to reaction with acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, the compound primarily responsible for the odor and taste of vinegar. Acetic acid reacts with potassium carbonate according to the following equation:

$$
2 \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{K}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{~K}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)
$$

The bubbling was due to the production of $\mathrm{CO}_{2}$.
The test of vinegar with potassium carbonate is one type of quantitative analysis-the determination of the
amount or concentration of a substance in a sample. In the analysis of vinegar, the concentration of the solute (acetic acid) was determined from the amount of reactant that combined with the solute present in a known volume of the solution. In other types of chemical analyses, the amount of a substance present in a sample is determined by measuring the amount of product that results.


743. Titration -
The described approach to measuring vinegar strength was an early version of the analytical technique known as titration analysis. A typical titration analysis involves the use of a buret (Figure 7.16) to make incremental additions of a solution containing a known concentration of some substance (the titrant) to a sample solution containing the substance whose concentration is to be measured (the analyte). The titrant and analyte undergo a chemical reaction of known stoichiometry, and so measuring the volume of titrant solution required for complete reaction with the analyte (the equivalence point of the titration) allows calculation of the analyte concentration. The equivalence point of a titration may be detected visually if a distinct change in the appearance of the sample solution accompanies the completion of the reaction. The halt of bubble formation in the classic vinegar analysis is one such example, though, more commonly, special dyes called indicators are added to the sample solutions to impart a change in color at or very near the equivalence point of the titration. Equivalence points may also be detected by measuring some solution property that changes in a predictable way during the course of the titration. Regardless of the approach taken to detect a titration's equivalence point, the volume of titrant actually measured is called the end point. Properly designed titration methods typically ensure that the difference between the equivalence and end points is negligible. Though any type of chemical reaction may serve as the basis for a titration analysis, the three described in this chapter (precipitation, acid-base, and redox) are most common. Additional details regarding titration analysis are provided in the chapter on acid-base equilibria.

FIGURE 7.16 (a) A student fills a buret in preparation for a titration analysis. (b) A typical buret permits volume measurements to the nearest 0.01 mL . (credit a: modification of work by Mark Blaser and Matt Evans; credit b: modification of work by Mark Blaser and Matt Evans)


745. Titration Analysis -
The end point in a titration of a $50.00-\mathrm{mL}$ sample of aqueous HCl was reached by addition of 35.23 mL of 0.250 M NaOH titrant. The titration reaction is:

$$
\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
$$

What is the molarity of the HCl ?


746. Solution -
As for all reaction stoichiometry calculations, the key issue is the relation between the molar amounts of the chemical species of interest as depicted in the balanced chemical equation. The approach outlined in previous modules of this chapter is followed, with additional considerations required, since the amounts of reactants provided and requested are expressed as solution concentrations.

For this exercise, the calculation will follow the following outlined steps:

The molar amount of HCl is calculated to be:

$$
35.23 \mathrm{~mL} \mathrm{NaOHI} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} \times \frac{0.250 \mathrm{~mol} \mathrm{NaOH}}{1 \mathrm{~L}} \times \frac{1 \mathrm{~mol} \mathrm{HCl}}{1 \mathrm{~mol} \mathrm{NaOH}}=8.81 \times 10^{-3} \mathrm{~mol} \mathrm{HCl}
$$

Using the provided volume of HCl solution and the definition of molarity, the HCl concentration is:

$$
\begin{aligned}
M & =\frac{\text { mol HCl }}{\mathrm{L} \text { solution }} \\
M & =\frac{8.81 \times 10^{-3} \mathrm{~mol} \mathrm{HCl}}{50.00 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}} \\
M & =0.176 M
\end{aligned}
$$

Note: For these types of titration calculations, it is convenient to recognize that solution molarity is also equal to the number of millimoles of solute per milliliter of solution:

$$
M=\frac{\text { mol solute }}{\mathrm{L} \text { solution }} \times \frac{\frac{10^{3} \mathrm{mmol}}{\mathrm{~mol}}}{\frac{10^{3} \mathrm{~mL}}{\mathrm{~L}}}=\frac{\mathrm{mmol} \text { solute }}{\mathrm{mL} \text { solution }}
$$


746. Solution -
Using this version of the molarity unit will shorten the calculation by eliminating two conversion factors:


747. Check Your Learning -
A $20.00-\mathrm{mL}$ sample of aqueous oxalic acid, $\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$, was titrated with a $0.09113-M$ solution of potassium permanganate, $\mathrm{KMnO}_{4}$ (see net ionic equation below).

$$
2 \mathrm{MnO}_{4}^{-}(a q)+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 10 \mathrm{CO}_{2}(g)+2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)
$$

A volume of 23.24 mL was required to reach the end point. What is the oxalic acid molarity?


748. Answer: -
0.2648 M


749. Gravimetric Analysis -
A gravimetric analysis is one in which a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from the other components of the sample. Mass measurements of the sample, the isolated analyte, or some other component of the analysis system, used along with the known stoichiometry of the compounds involved, permit calculation of the analyte concentration. Gravimetric methods were the first techniques used for quantitative chemical analysis, and they remain important tools in the modern chemistry laboratory.

The required change of state in a gravimetric analysis may be achieved by various physical and chemical processes. For example, the moisture (water) content of a sample is routinely determined by measuring the mass of a sample before and after it is subjected to a controlled heating process that evaporates the water. Also common are gravimetric techniques in which the analyte is subjected to a precipitation reaction of the sort described earlier in this chapter. The precipitate is typically isolated from the reaction mixture by filtration, carefully dried, and then weighed (Figure 7.17). The mass of the precipitate may then be used, along with relevant stoichiometric relationships, to calculate analyte concentration.

FIGURE 7.17 Precipitate may be removed from a reaction mixture by filtration.


751. Gravimetric Analysis -
A $0.4550-\mathrm{g}$ solid mixture containing $\mathrm{MgSO}_{4}$ is dissolved in water and treated with an excess of $\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$, resulting in the precipitation of 0.6168 g of $\mathrm{BaSO}_{4}$.

$$
\mathrm{MgSO}_{4}(a q)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{BaSO}_{4}(s)+\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q)
$$

What is the concentration (mass percent) of $\mathrm{MgSO}_{4}$ in the mixture?


752. Solution -
The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of $\mathrm{BaSO}_{4}$ and $\mathrm{MgSO}_{4}$ through their stoichiometric factor. Once the mass of $\mathrm{MgSO}_{4}$ is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.

The mass of $\mathrm{MgSO}_{4}$ that would yield the provided precipitate mass is

The concentration of $\mathrm{MgSO}_{4}$ in the sample mixture is then calculated to be

$$
\begin{aligned}
& \text { percent } \mathrm{MgSO}_{4}=\frac{\text { mass } \mathrm{MgSO}_{4}}{\text { mass sample }} \times 100 \% \\
& \frac{0.3181 \mathrm{~g}}{0.4550 \mathrm{~g}} \times 100 \%=69.91 \%
\end{aligned}
$$


753. Check Your Learning -
What is the percent of chloride ion in a sample if 1.1324 g of the sample produces 1.0881 g of AgCl when treated with excess $\mathrm{Ag}^{+}$?

$$
\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)
$$


754. Answer: -
23.76\%

The elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product (Figure 7.18). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.

FIGURE 7.18 This schematic diagram illustrates the basic components of a combustion analysis device for determining the carbon and hydrogen content of a sample.


755. Combustion Analysis -
Polyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a $0.00126-\mathrm{g}$ sample of polyethylene yields 0.00394 g of $\mathrm{CO}_{2}$ and 0.00161 g of $\mathrm{H}_{2} \mathrm{O}$. What is the empirical formula of polyethylene?


756. Solution -
The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:

$$
\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}(s)+\operatorname{excess} \mathrm{O}_{2}(g) \longrightarrow x \mathrm{CO}_{2}(g)+\frac{y}{2} \mathrm{H}_{2} \mathrm{O}(g)
$$

Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts $x$ and $y$ are needed.

First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:

$$
\begin{aligned}
& \text { mol C }=0.00394 \mathrm{~g} \mathrm{CO}_{2} \times \frac{1 \mathrm{~mol} \mathrm{CO}_{2}}{44.01 \mathrm{~g}} \times \frac{1 \mathrm{~mol} \mathrm{C}_{2}}{1 \mathrm{~mol} \mathrm{CO}_{2}}=8.95 \times 10^{-5} \mathrm{~mol} \mathrm{C} \\
& \mathrm{~mol} \mathrm{H}
\end{aligned}=0.00161 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \times \frac{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{18.02 \mathrm{~g}} \times \frac{2 \mathrm{~mol} \mathrm{H}^{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}=1.79 \times 10^{-4} \mathrm{~mol} \mathrm{H}}{}
$$


756. Solution -
The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H -to- C molar ratio is

$$
\frac{\mathrm{mol} \mathrm{H}}{\mathrm{~mol} \mathrm{C}}=\frac{1.79 \times 10^{-4} \mathrm{~mol} \mathrm{H}}{8.95 \times 10^{-5} \mathrm{~mol} \mathrm{C}}=\frac{2 \mathrm{~mol} \mathrm{H}}{1 \mathrm{~mol} \mathrm{C}}
$$

and the empirical formula for polyethylene is $\mathrm{CH}_{2}$.


757. Check Your Learning -
A $0.00215-\mathrm{g}$ sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of $\mathrm{CO}_{2}$ and 0.00148 g of $\mathrm{H}_{2} \mathrm{O}$ in a combustion analysis. What is the empirical formula for polystyrene?


758. Answer: -
CH


759. Key Terms -
acid substance that produces $\mathrm{H}_{3} \mathrm{O}^{+}$when dissolved in water
acid-base reaction reaction involving the transfer of a hydrogen ion between reactant species
actual yield amount of product formed in a reaction
analyte chemical species of interest
balanced equation chemical equation with equal numbers of atoms for each element in the reactant and product
base substance that produces $\mathrm{OH}^{-}$when dissolved in water
buret device used for the precise delivery of variable liquid volumes, such as in a titration analysis
chemical equation symbolic representation of a chemical reaction
coefficient number placed in front of symbols or formulas in a chemical equation to indicate their relative amount
combustion analysis gravimetric technique used to determine the elemental composition of a compound via the collection and weighing of its gaseous combustion products
combustion reaction vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light
complete ionic equation chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions
end point measured volume of titrant solution that yields the change in sample solution appearance or other property expected for stoichiometric equivalence (see equivalence point)
equivalence point volume of titrant solution required to react completely with the analyte in a titration analysis; provides a stoichiometric amount of titrant for the sample's analyte according to the titration reaction
excess reactant reactant present in an amount greater than required by the reaction stoichiometry
gravimetric analysis quantitative chemical analysis method involving the separation of an analyte from a sample by a physical or chemical process and subsequent mass measurements of the analyte, reaction product, and/or sample
half-reaction an equation that shows whether each reactant loses or gains electrons in a reaction.
indicator substance added to the sample in a
titration analysis to permit visual detection of the end point
insoluble of relatively low solubility; dissolving only to a slight extent
limiting reactant reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated
molecular equation chemical equation in which all reactants and products are represented as neutral substances


759. Key Terms -
limiting reactant reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated
molecular equation chemical equation in which all reactants and products are represented as neutral substances
net ionic equation chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)
neutralization reaction reaction between an acid and a base to produce salt and water
oxidation process in which an element's oxidation number is increased by loss of electrons
oxidation number (also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic
oxidation-reduction reaction (also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements
oxidizing agent (also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced
percent yield measure of the efficiency of a reaction, expressed as a percentage of the theoretical yield
precipitate insoluble product that forms from reaction of soluble reactants
precipitation reaction reaction that produces one or more insoluble products; when reactants are ionic compounds, sometimes called doubledisplacement or metathesis
product substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation
quantitative analysis the determination of the amount or concentration of a substance in a sample
reactant substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation
reducing agent (also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized
reduction process in which an element's oxidation number is decreased by gain of electrons
salt ionic compound that can be formed by the
reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide
single-displacement reaction (also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species
solubility the extent to which a substance may be dissolved in water, or any solvent
soluble of relatively high solubility; dissolving to a relatively large extent
spectator ion ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality


759. Key Terms -
soluble of relatively high solubility; dissolving to a relatively large extent
spectator ion ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality
stoichiometric factor ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products
stoichiometry relationships between the amounts


760. Key Equations -
percent yield $=\left(\frac{\text { actual yield }}{\text { theoretical yield }}\right) \times 100$


761. Summary - 761.1. Writing and Balancing Chemical Equations
Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products. Chemical reactions in aqueous solution that involve ionic reactants or products may be represented more realistically by complete ionic equations and, more succinctly, by net ionic equations.


761. Summary - 761.2. Classifying Chemical Reactions
Chemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: precipitation, acid-base, and oxidationreduction (redox). Precipitation reactions involve the formation of one or more insoluble products. Acid-base reactions involve the transfer of hydrogen ions between reactants. Redox reactions involve a change in oxidation number for one or more reactant elements. Writing balanced equations for
of reactants and products of a chemical reaction
strong acid acid that reacts completely when dissolved in water to yield hydronium ions
strong base base that reacts completely when dissolved in water to yield hydroxide ions
theoretical yield amount of product that may be produced from a given amount of reactant(s) according to the reaction stoichiometry
titrant solution containing a known concentration of substance that will react with the analyte in a titration analysis
titration analysis quantitative chemical analysis method that involves measuring the volume of a reactant solution required to completely react with the analyte in a sample
weak acid acid that reacts only to a slight extent when dissolved in water to yield hydronium ions
weak base base that reacts only to a slight extent when dissolved in water to yield hydroxide ions
some redox reactions that occur in aqueous solutions is simplified by using a systematic approach called the half-reaction method.


761. Summary - 761.3. Reaction Stoichiometry
A balanced chemical equation may be used to describe a reaction's stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.


761. Summary - 761.4. Reaction Yields
When reactions are carried out using less-thanstoichiometric quantities of reactants, the amount of product generated will be determined by the limiting reactant. The amount of product generated by a chemical reaction is its actual yield. This yield is often less than the amount of product predicted by the stoichiometry of the balanced chemical equation representing the reaction (its theoretical yield). The extent to which a reaction generates the theoretical amount of product is expressed as its percent yield.


761. Summary - 761.5. Quantitative Chemical Analysis
The stoichiometry of chemical reactions may serve as the basis for quantitative chemical analysis methods. Titrations involve measuring the volume of a titrant solution required to completely react with a sample solution. This volume is then used to calculate the concentration of analyte in the sample using the stoichiometry of the titration reaction. Gravimetric analysis involves separating the analyte
from the sample by a physical or chemical process, determining its mass, and then calculating its concentration in the sample based on the stoichiometry of the relevant process. Combustion analysis is a gravimetric method used to determine the elemental composition of a compound by collecting and weighing the gaseous products of its combustion.


762. Exercises - 762.1. Writing and Balancing Chemical Equations
1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?
2. Consider molecular, complete ionic, and net ionic equations.
(a) What is the difference between these types of equations?
(b) In what circumstance would the complete and net ionic equations for a reaction be identical?
3. Balance the following equations:
(a) $\mathrm{PCl}_{5}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{POCl}_{3}(l)+\mathrm{HCl}(a q)$
(b) $\mathrm{Cu}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NO}(g)$
(c) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{HI}(s)$
(d) $\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$
(e) $\mathrm{Na}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)$
(f) $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(s) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
(g) $\mathrm{P}_{4}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{3}(l)$
(h) $\mathrm{PtCl}_{4}(s) \longrightarrow \mathrm{Pt}(s)+\mathrm{Cl}_{2}(g)$
4. Balance the following equations:


762. Exercises - 762.1. Writing and Balancing Chemical Equations
(h) $\mathrm{PtCl}_{4}(s) \longrightarrow \mathrm{Pt}(s)+\mathrm{Cl}_{2}(g)$
4. Balance the following equations:
(a) $\mathrm{Ag}(s)+\mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Ag}_{2} \mathrm{~S}(s)+\mathrm{H}_{2} \mathrm{O}(l)$
(b) $\mathrm{P}_{4}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)$
(c) $\mathrm{Pb}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Pb}(\mathrm{OH})_{2}(s)$
(d) $\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g)$
(e) $\mathrm{Sc}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(l) \longrightarrow \mathrm{Sc}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s)$
(f) $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}(a q)$
(g) $\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(s)+\mathrm{H}_{2}(g)$
(h) $\mathrm{TiCl}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{TiO}_{2}(s)+\mathrm{HCl}(g)$
5. Write a balanced molecular equation describing each of the following chemical reactions.


762. Exercises - 762.1. Writing and Balancing Chemical Equations
5. Write a balanced molecular equation describing each of the following chemical reactions.
(a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
(b) Gaseous butane, $\mathrm{C}_{4} \mathrm{H}_{10}$, reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.
(c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
(d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.
6. Write a balanced equation describing each of the following chemical reactions.
(a) Solid potassium chlorate, $\mathrm{KClO}_{3}$, decomposes to form solid potassium chloride and diatomic oxygen gas.
(b) Solid aluminum metal reacts with solid diatomic iodine to form solid $\mathrm{Al}_{2} \mathrm{I}_{6}$.
(c) When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.
(d) Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.
7. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.
(a) Write the formulas of barium nitrate and potassium chlorate.
(b) The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.
(c) The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.
(d) Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains $\mathrm{Fe}^{3+}$ ions.)
8. Fill in the blank with a single chemical formula for a covalent compound that will balance the equation:


762. Exercises - 762.1. Writing and Balancing Chemical Equations
9. Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).
(a) Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.
(b) The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write complete and net ionic equations for this reaction.
10. A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process.
(a) The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.
(b) The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.
(c) Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.
(d) The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.
(e) Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.
11. From the balanced molecular equations, write the complete ionic and net ionic equations for the following:
(a) $\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow 2 \mathrm{KOH}(a q)+\mathrm{BaC}_{2} \mathrm{O}_{4}(s)$
(b) $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+2 \mathrm{HNO}_{3}(a q)$


762. Exercises - 762.1. Writing and Balancing Chemical Equations
(c) $\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$


762. Exercises - 762.2. Classifying Chemical Reactions
12. Use the following equations to answer the next four questions:
i. $\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$
ii. $\mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Ag}^{+}(a q)+\mathrm{NO}_{3}{ }^{-}(a q) \longrightarrow \mathrm{AgCl}(s)+\mathrm{Na}^{+}(a q)+\mathrm{NO}_{3}{ }^{-}(a q)$
iii. $\mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
iv. $2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$
v. $\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$
(a) Which equation describes a physical change?
(b) Which equation identifies the reactants and products of a combustion reaction?
(c) Which equation is not balanced?
(d) Which is a net ionic equation?
13. Indicate what type, or types, of reaction each of the following represents:
(a) $\mathrm{Ca}(s)+\mathrm{Br}_{2}(l) \longrightarrow \mathrm{CaBr}_{2}(s)$
(b) $\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{HBr}(a q) \longrightarrow \mathrm{CaBr}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$
(c) $\mathrm{C}_{6} \mathrm{H}_{12}(l)+9 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$
14. Indicate what type, or types, of reaction each of the following represents:


762. Exercises - 762.2. Classifying Chemical Reactions
14. Indicate what type, or types, of reaction each of the following represents:
(a) $\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$
(b) $2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$
(c) $\mathrm{Al}(\mathrm{OH})_{3}(a q)+3 \mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)$
(d) $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+2 \mathrm{HNO}_{3}(a q)$
15. Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer.
16. Determine the oxidation states of the elements in the following compounds:
(a) NaI
(b) $\mathrm{GdCl}_{3}$
(c) $\mathrm{LiNO}_{3}$
(d) $\mathrm{H}_{2} \mathrm{Se}$
(e) $\mathrm{Mg}_{2} \mathrm{Si}$
(f) $\mathrm{RbO}_{2}$, rubidium superoxide
(g) HF
17. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.
(a) $\mathrm{H}_{3} \mathrm{PO}_{4}$
(b) $\mathrm{Al}(\mathrm{OH})_{3}$
(c) $\mathrm{SeO}_{2}$
(d) $\mathrm{KNO}_{2}$
(e) $\mathrm{In}_{2} \mathrm{~S}_{3}$
(f) $\mathrm{P}_{4} \mathrm{O}_{6}$


762. Exercises - 762.2. Classifying Chemical Reactions
(d) $\mathrm{KNO}_{2}$
(e) $\mathrm{In}_{2} \mathrm{~S}_{3}$
(f) $\mathrm{P}_{4} \mathrm{O}_{6}$
18. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.
(a) $\mathrm{H}_{2} \mathrm{SO}_{4}$
(b) $\mathrm{Ca}(\mathrm{OH})_{2}$
(c) BrOH
(d) $\mathrm{ClNO}_{2}$
(e) $\mathrm{TiCl}_{4}$
(f) NaH
19. Classify the following as acid-base reactions or oxidation-reduction reactions:
(a) $\mathrm{Na}_{2} \mathrm{~S}(a q)+2 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)$
(b) $2 \mathrm{Na}(s)+2 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{H}_{2}(g)$
(c) $\mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{MgCl}_{2}(s)$
(d) $\mathrm{MgO}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$
(e) $\mathrm{K}_{3} \mathrm{P}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(s)$
(f) $3 \mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)$
20. Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations:


762. Exercises - 762.2. Classifying Chemical Reactions
20. Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations:
(a) $\mathrm{Mg}(s)+\mathrm{NiCl}_{2}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{Ni}(s)$
(b) $\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)$
(c) $\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$
(d) $\mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g)$
(e) $2 \mathrm{~K}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(s)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{K}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}(s)+2 \mathrm{KI}(s)$
(f) $3 \mathrm{Cu}(s)+8 \mathrm{HNO}_{3}(a q) \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$
21. Complete and balance the following acid-base equations:
(a) HCl gas reacts with solid $\mathrm{Ca}(\mathrm{OH})_{2}(s)$.
(b) A solution of $\mathrm{Sr}(\mathrm{OH})_{2}$ is added to a solution of $\mathrm{HNO}_{3}$.
22. Complete and balance the following acid-base equations:
(a) A solution of $\mathrm{HClO}_{4}$ is added to a solution of LiOH .
(b) Aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$ reacts with NaOH .


762. Exercises - 762.2. Classifying Chemical Reactions
(a) A solution of $\mathrm{HClO}_{4}$ is added to a solution of LiOH .
(b) Aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$ reacts with NaOH .
(c) $\mathrm{Ba}(\mathrm{OH})_{2}$ reacts with HF gas.
23. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.
(a) $\mathrm{Al}(s)+\mathrm{F}_{2}(g) \longrightarrow$
(b) $\mathrm{Al}(s)+\mathrm{CuBr}_{2}(a q) \longrightarrow$ (single displacement)
(c) $\mathrm{P}_{4}(s)+\mathrm{O}_{2}(g) \longrightarrow$
(d) $\mathrm{Ca}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ (products are a strong base and a diatomic gas)
24. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.
(a) $\mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$
(b) $\mathrm{Ba}(s)+\mathrm{HBr}(a q) \longrightarrow$
(c) $\operatorname{Sn}(s)+\mathrm{I}_{2}(s) \longrightarrow$
25. Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used.
(a) $\mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow$
(b) $\mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ (assume an excess of water and that the product dissolves)
(c) $\mathrm{SrO}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow$


762. Exercises - 762.2. Classifying Chemical Reactions
(c) $\mathrm{SrO}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow$
26. When heated to $700-800^{\circ} \mathrm{C}$, diamonds, which are pure carbon, are oxidized by atmospheric oxygen. (They burn!) Write the balanced equation for this reaction.
27. The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?
28. Write the molecular, total ionic, and net ionic equations for the following reactions:
(a) $\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \longrightarrow$
(b) $\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{CaCl}_{2}(a q) \longrightarrow$
29. Great Lakes Chemical Company produces bromine, $\mathrm{Br}_{2}$, from bromide salts such as NaBr , in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with $\mathrm{Cl}_{2}$.
30. In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white solid, but in these experiments it often looks gray, due to small amounts of $\mathrm{Mg}_{3} \mathrm{~N}_{2}$, a compound formed as some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction.
31. Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of $\mathrm{CO}_{2}$. (Hint: Water is one of the products.)


762. Exercises - 762.2. Classifying Chemical Reactions
32. Calcium propionate is sometimes added to bread to retard spoilage. This compound can be prepared by the reaction of calcium carbonate, $\mathrm{CaCO}_{3}$, with propionic acid, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}$, which has properties similar to those of acetic acid. Write the balanced equation for the formation of calcium propionate.
33. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:
(a) $\mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow$
(b) $\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow$
34. Copper(II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper(II) oxide. The gaseous product then reacts with liquid water to produce liquid dihydrogen sulfate as the only product. Write the two equations which represent these reactions.
35. Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.
(a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)
(b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction
(c) gaseous $\mathrm{H}_{2} \mathrm{~S}$ from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)


762. Exercises - 762.2. Classifying Chemical Reactions
36. Calcium cyclamate $\mathrm{Ca}\left(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{NHSO}_{3}\right)_{2}$ is an artificial sweetener used in many countries around the world but is banned in the United States. It can be purified industrially by converting it to the barium salt through reaction of the acid $\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{NHSO}_{3} \mathrm{H}$ with barium carbonate, treatment with sulfuric acid (barium sulfate is very insoluble), and then neutralization with calcium hydroxide. Write the balanced equations for these reactions.
37. Complete and balance each of the following half-reactions (steps $2-5$ in half-reaction method):
(a) $\mathrm{Sn}^{4+}(a q) \longrightarrow \mathrm{Sn}^{2+}(a q)$
(b) $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) \longrightarrow \mathrm{Ag}(s)+\mathrm{NH}_{3}(a q)$
(c) $\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) \longrightarrow \mathrm{Hg}(l)+\mathrm{Cl}^{-}(a q)$
(d) $\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{O}_{2}(g)$ (in acidic solution)
(e) $\mathrm{IO}_{3}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s)$ (in basic solution)
(f) $\mathrm{SO}_{3}{ }^{2-}(a q) \longrightarrow \mathrm{SO}_{4}{ }^{2-}(a q)$ (in acidic solution)
(g) $\mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)$ (in acidic solution)
(h) $\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}{ }^{-}(a q)$ (in basic solution)
38. Complete and balance each of the following half-reactions (steps $2-5$ in half-reaction method):


762. Exercises - 762.2. Classifying Chemical Reactions
38. Complete and balance each of the following half-reactions (steps $2-5$ in half-reaction method):
(a) $\mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)$
(b) $\mathrm{Hg}(l)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{HgBr}_{4}{ }^{2-}(a q)$
(c) $\mathrm{ZnS}(s) \longrightarrow \mathrm{Zn}(s)+\mathrm{S}^{2-}(a q)$
(d) $\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}$ (l (in basic solution)
(e) $\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)$ (in acidic solution)
(f) $\mathrm{NO}_{3}{ }^{-}(a q) \longrightarrow \mathrm{HNO}_{2}(a q)$ (in acidic solution)
(g) $\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{MnO}_{4}{ }^{-}(a q)$ (in basic solution)
(h) $\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}{ }^{-}(a q)$ (in acidic solution)
39. Balance each of the following equations according to the half-reaction method:
(a) $\mathrm{Sn}^{2+}(a q)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q)+\mathrm{Cu}^{+}(a q)$
(b) $\mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{Hg}_{2}{ }^{2+}(a q) \longrightarrow \mathrm{Hg}(l)+\mathrm{S}(s)$ (in acid)
(c) $\mathrm{CN}^{-}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{CNO}^{-}(a q)+\mathrm{Cl}^{-}(a q)$ (in acid)


762. Exercises - 762.2. Classifying Chemical Reactions
(d) $\mathrm{Fe}^{2+}(a q)+\mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)$
(e) $\mathrm{HBrO}(a q) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{O}_{2}(g)$ (in acid)
40. Balance each of the following equations according to the half-reaction method:
(a) $\mathrm{Zn}(s)+\mathrm{NO}_{3}{ }^{-}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{N}_{2}(g)$ (in acid)
(b) $\mathrm{Zn}(s)+\mathrm{NO}_{3}{ }^{-}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{NH}_{3}(a q)$ (in base)
(c) $\mathrm{CuS}(s)+\mathrm{NO}_{3}{ }^{-}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{S}(s)+\mathrm{NO}(g)$ (in acid)
(d) $\mathrm{NH}_{3}(a q)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)$ (gas phase)
(e) $\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{O}_{2}(g)$ (in acid)
(f) $\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}_{3}{ }^{-}(a q)+\mathrm{NO}_{2}{ }^{-}(a q)$ (in base)
(g) $\mathrm{Fe}^{3+}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{I}_{2}(a q)$
41. Balance each of the following equations according to the half-reaction method:


762. Exercises - 762.2. Classifying Chemical Reactions
41. Balance each of the following equations according to the half-reaction method:
(a) $\mathrm{MnO}_{4}{ }^{-}(a q)+\mathrm{NO}_{2}{ }^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{NO}_{3}{ }^{-}(a q)$ (in base)
(b) $\mathrm{MnO}_{4}{ }^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}{ }^{-}(a q)+\mathrm{MnO}_{2}(s)$ (in base)
(c) $\mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q)$ (in acid)


762. Exercises - 762.3. Reaction Stoichiometry
42. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:
(a) The number of moles and the mass of chlorine, $\mathrm{Cl}_{2}$, required to react with 10.0 g of sodium metal, Na , to produce sodium chloride, NaCl .
(b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.
(c) The number of moles and the mass of sodium nitrate, $\mathrm{NaNO}_{3}$, required to produce 128 g of oxygen. ( $\mathrm{NaNO}_{2}$ is the other product.)
(d) The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 kg of carbon in an excess of oxygen.
(e) The number of moles and the mass of copper(II) carbonate needed to produce 1.500 kg of copper(II) oxide. ( $\mathrm{CO}_{2}$ is the other product.)
(f)

The number of moles and the mass of
formed by the reaction of 12.85 g of


762. Exercises - 762.3. Reaction Stoichiometry
The number of moles and the mass of
formed by the reaction of 12.85 g of

with an excess of $\mathrm{Br}_{2}$.
43. Determine the number of moles and the mass requested for each reaction in Exercise 7.42.
44. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:
(a) The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce $\mathrm{MgCl}_{2}$ and $\mathrm{H}_{2}$.
(b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of $\operatorname{silver(I)~oxide.~}$
(c) The number of moles and the mass of magnesium carbonate, $\mathrm{MgCO}_{3}$, required to produce 283 g of carbon dioxide. (MgO is the other product.)
(d) The number of moles and the mass of water formed by the combustion of 20.0 kg of acetylene, $\mathrm{C}_{2} \mathrm{H}_{2}$, in an excess of oxygen.
(e) The number of moles and the mass of barium peroxide, $\mathrm{BaO}_{2}$, needed to produce 2.500 kg of barium oxide, $\mathrm{BaO}\left(\mathrm{O}_{2}\right.$ is the other product.)
(f)

The number of moles and the mass of


762. Exercises - 762.3. Reaction Stoichiometry
45. Determine the number of moles and the mass requested for each reaction in Exercise 7.44.
46. $\mathrm{H}_{2}$ is produced by the reaction of 118.5 mL of a $0.8775-\mathrm{M}$ solution of $\mathrm{H}_{3} \mathrm{PO}_{4}$ according to the following equation: $2 \mathrm{Cr}+2 \mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow 3 \mathrm{H}_{2}+2 \mathrm{CrPO}_{4}$.
(a) Outline the steps necessary to determine the number of moles and mass of $\mathrm{H}_{2}$.
(b) Perform the calculations outlined.
47. Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: $2 \mathrm{Ga}+6 \mathrm{HCl} \longrightarrow 2 \mathrm{GaCl}_{3}+3 \mathrm{H}_{2}$.
(a) Outline the steps necessary to determine the number of moles and mass of gallium chloride.
(b) Perform the calculations outlined.
48. $\mathrm{I}_{2}$ is produced by the reaction of 0.4235 mol of $\mathrm{CuCl}_{2}$ according to the following equation: $2 \mathrm{CuCl}_{2}+4 \mathrm{KI} \longrightarrow 2 \mathrm{CuI}+4 \mathrm{KCl}+\mathrm{I}_{2}$.
(a) How many molecules of $\mathrm{I}_{2}$ are produced?
(b) What mass of $\mathrm{I}_{2}$ is produced?
49. Silver is often extracted from ores such as $\mathrm{K}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]$ and then recovered by the reaction $2 \mathrm{~K}\left\mathrm{Ag}(\mathrm{CN})_{2}\right+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{Zn}(\mathrm{CN})_{2}(a q)+2 \mathrm{KCN}(a q)$


762. Exercises - 762.3. Reaction Stoichiometry
(a) How many molecules of $\mathrm{Zn}(\mathrm{CN})_{2}$ are produced by the reaction of 35.27 g of $\mathrm{K}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]$ ?
(b) What mass of $\mathrm{Zn}(\mathrm{CN})_{2}$ is produced?
50. What mass of silver oxide, $\mathrm{Ag}_{2} \mathrm{O}$, is required to produce 25.0 g of silver sulfadiazine, $\mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{~N}_{4} \mathrm{SO}_{2}$, from the reaction of silver oxide and sulfadiazine?
$2 \mathrm{C}_{10} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{SO}_{2}+\mathrm{Ag}_{2} \mathrm{O} \longrightarrow 2 \mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{~N}_{4} \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}$
51. Carborundum is silicon carbide, SiC , a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, $\mathrm{SiO}_{2}$, with carbon at high temperature. Carbon monoxide, CO , is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much $\mathrm{SiO}_{2}$ is required to produce 3.00 kg of SiC .
52. Automotive air bags inflate when a sample of sodium azide, $\mathrm{NaN}_{3}$, is very rapidly decomposed.
$2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{~N}_{2}(g)$
What mass of sodium azide is required to produce $2.6 \mathrm{ft}^{3}(73.6 \mathrm{~L})$ of nitrogen gas with a density of $1.25 \mathrm{~g} /$ L?


762. Exercises - 762.3. Reaction Stoichiometry
53. Urea, $\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}$, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the $\mathrm{CO}_{2}$ produced by combustion of $1.00 \times 10^{3} \mathrm{~kg}$ of carbon followed by the reaction?
$\mathrm{CO}_{2}(g)+2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)$
54. In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ was quickly spread on the area and $\mathrm{CO}_{2}$ was released by the reaction. Was sufficient $\mathrm{Na}_{2} \mathrm{CO}_{3}$ used to neutralize all of the acid?
55. A compact car gets 37.5 miles per gallon on the highway. If gasoline contains $84.2 \%$ carbon by mass and has a density of $0.8205 \mathrm{~g} / \mathrm{mL}$, determine the mass of carbon dioxide produced during a 500-mile trip (3.785 liters per gallon).
56. What volume of 0.750 M hydrochloric acid solution can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with excess sulfuric acid?
$\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{HCl}(g)+\mathrm{NaHSO}_{4}(s)$


762. Exercises - 762.3. Reaction Stoichiometry
$\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{HCl}(g)+\mathrm{NaHSO}_{4}(s)$
57. What volume of a 0.2089 MKI solution contains enough KI to react exactly with the $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ in 43.88 mL of a 0.3842 M solution of $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ ? $2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+4 \mathrm{KI} \longrightarrow 2 \mathrm{CuI}+\mathrm{I}_{2}+4 \mathrm{KNO}_{3}$
58. A mordant is a substance that combines with a dye to produce a stable fixed color in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide.
$2 \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{Ca}(\mathrm{OH})_{2} \longrightarrow \mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{CO}_{2}\right)_{2}+2 \mathrm{H}_{2} \mathrm{O}$
What mass of $\mathrm{Ca}(\mathrm{OH})_{2}$ is required to react with the acetic acid in 25.0 mL of a solution having a density of $1.065 \mathrm{~g} / \mathrm{mL}$ and containing $58.0 \%$ acetic acid by mass?


762. Exercises - 762.3. Reaction Stoichiometry
59. The toxic pigment called white lead, $\mathrm{Pb}_{3}(\mathrm{OH})_{2}\left(\mathrm{CO}_{3}\right)_{2}$, has been replaced in white paints by rutile, $\mathrm{TiO}_{2}$. How much rutile (g) can be prepared from 379 g of an ore that contains $88.3 \%$ ilmenite $\left(\mathrm{FeTiO}_{3}\right)$ by mass? $2 \mathrm{FeTiO}_{3}+4 \mathrm{HCl}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{FeCl}_{3}+2 \mathrm{TiO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$


762. Exercises - 762.4. Reaction Yields
60. The following quantities are placed in a container: $1.5 \times 10^{24}$ atoms of hydrogen, 1.0 mol of sulfur, and 88.0 g of diatomic oxygen.
(a) What is the total mass in grams for the collection of all three elements?
(b) What is the total number of moles of atoms for the three elements?
(c) If the mixture of the three elements formed a compound with molecules that contain two hydrogen atoms, one sulfur atom, and four oxygen atoms, which substance is consumed first?
(d) How many atoms of each remaining element would remain unreacted in the change described in (c)?
61. What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine?
62. Which of the postulates of Dalton's atomic theory explains why we can calculate a theoretical yield for a chemical reaction?
63. A student isolated 25 g of a compound following a procedure that would theoretically yield 81 g . What was his percent yield?
64. A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. What is the percent yield for this reaction?
$\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(s)$
65. Freon-12, $\mathrm{CCl}_{2} \mathrm{~F}_{2}$, is prepared from $\mathrm{CCl}_{4}$ by reaction with HF . The other product of this reaction is HCl . Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of $\mathrm{CCl}_{2} \mathrm{~F}_{2}$ from 32.9 g of $\mathrm{CCl}_{4}$. Freon-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield.


762. Exercises - 762.4. Reaction Yields
66. Citric acid, $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}$, a component of jams, jellies, and fruity soft drinks, is prepared industrially via fermentation of sucrose by the mold Aspergillus niger. The equation representing this reaction is $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}+4 \mathrm{H}_{2} \mathrm{O}$
What mass of citric acid is produced from exactly 1 metric ton $\left(1.000 \times 10^{3} \mathrm{~kg}\right)$ of sucrose if the yield is $92.30 \%$ ?
67. Toluene, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}$, is oxidized by air under carefully controlled conditions to benzoic acid, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}$, which is used to prepare the food preservative sodium benzoate, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}$. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid?
$2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}+2 \mathrm{H}_{2} \mathrm{O}$
68. In a laboratory experiment, the reaction of 3.0 mol of $\mathrm{H}_{2}$ with 2.0 mol of $\mathrm{I}_{2}$ produced 1.0 mol of HI . Determine the theoretical yield in grams and the percent yield for this reaction.


762. Exercises - 762.4. Reaction Yields
69. Outline the steps needed to solve the following problem, then do the calculations. Ether, $\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}$, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.
$2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}$
What is the percent yield of ether if $1.17 \mathrm{~L}(\mathrm{~d}=0.7134 \mathrm{~g} / \mathrm{mL})$ is isolated from the reaction of 1.500 L of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ $(\mathrm{d}=0.7894 \mathrm{~g} / \mathrm{mL})$ ?
70. Outline the steps needed to determine the limiting reactant when 30.0 g of propane, $\mathrm{C}_{3} \mathrm{H}_{8}$, is burned with 75.0 g of oxygen.


762. Exercises - 762.4. Reaction Yields
Determine the limiting reactant.
71. Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of $\mathrm{H}_{3} \mathrm{PO}_{4}$ react according to the following chemical equation.
$2 \mathrm{Cr}+2 \mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow 2 \mathrm{CrPO}_{4}+3 \mathrm{H}_{2}$
Determine the limiting reactant.
72. What is the limiting reactant when 1.50 g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation?
$\mathrm{Li}+\mathrm{N}_{2} \longrightarrow \mathrm{Li}_{3} \mathrm{~N}$
73. Uranium can be isolated from its ores by dissolving it as $\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}$, then separating it as solid $\mathrm{UO}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right) \cdot 3 \mathrm{H}_{2} \mathrm{O}$. Addition of 0.4031 g of sodium oxalate, $\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$, to a solution containing 1.481 g of uranyl nitrate, $\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}$, yields 1.073 g of solid $\mathrm{UO}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right) \cdot 3 \mathrm{H}_{2} \mathrm{O}$.


762. Exercises - 762.4. Reaction Yields
$\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{UO}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right) \cdot 3 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{NaNO}_{3}$
Determine the limiting reactant and the percent yield of this reaction.
74. How many molecules of $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}$ can be prepared from $15 \mathrm{C}_{2} \mathrm{H}_{4}$ molecules and $8 \mathrm{Cl}_{2}$ molecules?
75. How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms?