# 12.6: Mass-Volume and Volume-Mass Stoichiometry

Cars and many other vehicles have air bags in them. In case of a collision, a reaction is triggered so that the rapid decomposition of sodium azide produces nitrogen gas, filling the air bag. If too little sodium azide is used, the air bag will not fill completely and will not protect the person in the vehicle. Too much sodium azide could cause the formation of more gas that the bag can safely handle. If the bag breaks from the excess gas pressure, all protection is lost.

### Mass to Volume and Volume to Mass Problems

Chemical reactions frequently involve both solid substances whose mass can be measured as well as gases for which measuring the volume is more appropriate. Stoichiometry problems of this type are called either mass-volume or volume-mass problems.

$\text{mass of given} \rightarrow \text{moles of given} \rightarrow \text{moles of unknown} \rightarrow \text{volume of unknown}$

$\text{volume of given} \rightarrow \text{moles of given} \rightarrow \text{moles of unknown} \rightarrow \text{mass of unknown}$

Because both types of problems involve a conversion from either moles of gas to volume or vice-versa, we can use the molar volume of $$22.4 \: \text{L/mol}$$ provided that the conditions for the reaction are STP.

Example 12.6.1

Aluminum metal reacts rapidly with aqueous sulfuric acid to produce aqueous aluminum sulfate and hydrogen gas.

$2 \: \text{Al} \left( s \right) + 3 \ce{H_2SO_4} \left( aq \right) \rightarrow \ce{Al_2(SO_4)_3} \left( aq \right) + 3 \ce{H_2} \left( g \right)$

Determine the volume of hydrogen gas produced at STP when a $$2.00 \: \text{g}$$ piece of aluminum completely reacts.

Solution:

Step 1: List the known quantities and plan the problem.

Known

• Given: $$2.00 \: \text{g} \: \ce{Al}$$
• Molar mass $$\ce{Al} = 26.98 \: \text{g/mol}$$
• $$2 \: \text{mol} \: \ce{Al} = 3 \: \text{mol} \: \ce{H_2}$$

Unknown

• Volume $$\ce{H_2} = ?$$

The grams of aluminum will first be converted to moles. Then the mole ratio will be applied to convert to moles of hydrogen gas. Finally, the molar volume of a gas will be used to convert to liters of hydrogen.

$\text{g} \: \ce{Al} \rightarrow \text{mol} \: \ce{Al} \rightarrow \text{mol} \: \ce{H_2} \rightarrow \text{L} \: \ce{H_2}$

Step 2: Solve.

$2.00 \: \text{g} \: \ce{Al} \times \frac{1 \: \text{mol} \: \ce{Al}}{26.98 \: \text{g} \: \ce{Al}} \times \frac{3 \: \text{mol} \: \ce{H_2}}{2 \: \text{mol} \: \ce{Al}} \times \frac{22.4 \: \text{L} \: \ce{H_2}}{1 \: \text{mol} \: \ce{H_2}} = 2.49 \: \text{L} \: \ce{H_2}$

The volume result is in liters. For much smaller amounts, it may be convenient to convert to milliliters. The answer here has three significant figures. Because the molar volume is a measured quantity of $$22.4 \: \text{L/mol}$$, three is the maximum number of significant figures for this type of problem.

Example 12.6.2

Calcium oxide is used to remove sulfur dioxide generated in coal-burning power plants according to the following reaction.

$2 \ce{CaO} \left( s \right) + 2 \ce{SO_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{CaSO_4} \left( s \right)$

What mass of calcium oxide is required to react completely with $$1.4 \times 10^3 \: \text{L}$$ of sulfur dioxide?

Solution:

Step 1: List the known quantities and plan the problem.

Known

• Given: $$1.4 \times 10^3 \: \text{L} = \ce{SO_2}$$
• $$2 \: \text{mol} \: \ce{SO_2} = 2 \: \text{mol} \ce{CaO}$$
• Molar mass $$\ce{CaO} = 56.08 \: \text{g/mol}$$

Unknown

• Mass $$\ce{CaO} = ? \: \text{g}$$

The volume of $$\ce{SO_2}$$ will be converted to moles, followed by the mole ratio, and finally a conversion of moles of $$\ce{CaO}$$ to grams.

$\text{L} \: \ce{SO_2} \rightarrow \text{mol} \: \ce{SO_2} \rightarrow \text{mol} \: \ce{CaO} \rightarrow \text{g} \: \ce{CaO}$

Step 2: Solve.

$1.4 \times 10^3 \: \text{L} \: \ce{SO_2} \times \frac{1 \: \text{mol} \: \ce{SO_2}}{22.4 \: \text{L} \: \ce{SO_2}} \times \frac{2 \: \text{mol} \: \ce{CaO}}{2 \: \text{mol} \: \ce{SO_2}} \times \frac{56.08 \: \text{g} \: \ce{CaO}}{1 \: \text{mol} \: \ce{CaO}} = 3.5 \times 10^3 \: \text{g} \: \ce{CaO}$

The resultant mass could be reported as $$3.5 \: \text{kg}$$, with two significant figures. Even though the 2:2 mole ratio does not mathematically affect the problem, it is still necessary for unit conversion.