# 14.10: Gas Stoichiometry

The Haber cycle reaction of gaseous nitrogen and hydrogen to form ammonia is a critical step in the production of fertilizer from ammonia. It is important to have an excess of the starting materials so that a maximum yield of ammonia can be achieved. By knowing how much ammonia is needed for manufacture of a batch of fertilizer, the proper amounts of nitrogen and hydrogen gases can be incorporated into the process.

### Gas Stoichiometry

You have learned how to use molar volume to solve stoichiometry problems for chemical reactions involving one or more gases at STP. Now, we can use the ideal gas law to expand our treatment of chemical reactions to solve stoichiometry problems for reactions that occur at any temperature and pressure.

Example 14.10.1

What volume of carbon dioxide is produced by the combustion of $$25.21 \: \text{g}$$ of ethanol $$\left( \ce{C_2H_5OH} \right)$$ at $$54^\text{o} \text{C}$$ and $$728 \: \text{mm} \: \ce{Hg}$$?

Solution:

Before using the ideal gas law, it is necessary to write and balance the chemical equation. Recall that in most combustion reactions, the given substance reacts with $$\ce{O_2}$$ to form $$\ce{CO_2}$$ and $$\ce{H_2O}$$. Here is the balanced equation for the combustion of ethanol.

$\ce{C_2H_5OH} \left( l \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + 3 \ce{H_2O} \left( l \right)$

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

Known

• Mass $$\ce{C_2H_5OH} = 25.21 \: \text{g}$$
• Molar mass $$\ce{C_2H_5OH} = 46.08 \: \text{g/mol}$$
• $$P = 728 \: \text{mm} \: \ce{Hg}$$
• $$T = 54^\text{o} \text{C} = 327 \: \text{K}$$

Unknown

• Volume $$\ce{CO_2} = ? \: \text{L}$$

The number of moles of carbon dioxide gas is first calculated by stoichiometry. Then the ideal gas law is used to calculate the volume of $$\ce{CO_2}$$ produced.

Step 2: Solve.

$25.21 \: \text{g} \: \ce{C_2H_5OH} \times \frac{1 \: \text{mol} \: \ce{C_2H_5OH}}{46.08 \: \text{g} \: \ce{C_2H_5OH}} \times \frac{2 \: \text{mol} \: \ce{CO_2}}{1 \: \text{mol} \: \ce{C_2H_5OH}} = 1.094 \: \text{mol} \: \ce{CO_2}$

The moles of carbon dioxide $$\left( n \right)$$ is now substituted into $$PV = nRT$$ to solve for the volume.

$V = \frac{nRT}{P} = \frac{1.094 \: \text{mol} \times 62.36 \: \text{L} \cdot \text{mm} \: \text{Hg}/\text{K} \cdot \text{mol} \times 327 \: \text{K}}{728 \: \text{mm} \: \ce{Hg}} = 30.6 \: \text{L}$

The mass of ethanol is slightly more than one half mole, meaning that the mole ratio results in slightly more than one mole of carbon dioxide being produced. Because of the elevated temperature and reduced pressure compared to STP, the resulting volume is larger than $$22.4 \: \text{L}$$.