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# Gas Stoichiometry

## The ideal gas law is used to balance equations involving gases

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Gas Stoichiometry

#### How is fertilizer produced?

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.

#### Sample Problem: Gas Stoichiometry and the Ideal Gas Law

What volume of carbon dioxide is produced by the combustion of 25.21 g of ethanol (C2H5OH) at 54°C and 728 mmHg? Assume the gas is ideal.

Before using the ideal gas law, it is necessary to write and balance the chemical equation. Recall that most combustion reactions, the given substance reacts with O2 to form CO2 and H2O. Here is the balanced equation for the combustion of ethanol.

C2H5OH(l)+3O2(g)2CO2(g)+3H2O(l)\begin{align*}\text{C}_2\text{H}_5\text{OH}(l)+3\text{O}_2(g) \rightarrow 2\text{CO}_2(g)+3\text{H}_2\text{O}(l)\end{align*}

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

Known

• mass C2H5OH=25.21 g\begin{align*}\text{mass} \ \text{C}_2\text{H}_5\text{OH}=25.21 \text{ g}\end{align*}
• molar mass C2H5OH=46.08 g/mol\begin{align*}\text{molar mass} \ \text{C}_2\text{H}_5\text{OH}=46.08 \text{ g/mol}\end{align*}
• P=728 mmHg\begin{align*}P=728 \text{ mmHg}\end{align*}
• T=54C=327 K\begin{align*}T=54^\circ \text{C}=327 \text{ K}\end{align*}

Unknown

• Volume CO2=? L\begin{align*}\text{Volume} \ \text{CO}_2=? \text{ L}\end{align*}

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 CO2 produced.

Step 2: Solve.

25.21 g C2H5OH×1 mol C2H5OH46.08 g C2H5OH×2 mol CO21 mol C2H5OH=1.094 mol C2H5OH\begin{align*}25.21 \text{ g } \text{C}_2\text{H}_5\text{OH} \times \frac{1 \text{ mol } \text{C}_2\text{H}_5\text{OH}}{46.08 \text{ g } \text{C}_2\text{H}_5\text{OH}} \times \frac{2 \text{ mol } \text{CO}_2}{1 \text{ mol } \text{C}_2\text{H}_5\text{OH}}=1.094 \text{ mol } \text{C}_2\text{H}_5\text{OH}\end{align*}

The moles of ethanol (n)\begin{align*}(n)\end{align*} is now substituted into PV=nRT\begin{align*}PV=nRT\end{align*} to solve for the volume.

V=nRTP=1.094 mol×62.36 LmmHg/Kmol×327 K728 mmHg=30.6 L\begin{align*}V=\frac{nRT}{P}=\frac{1.094 \text{ mol} \times 62.36 \text{ L} \cdot \text{mmHg/K} \cdot \text{mol} \times 327 \text{ K}}{728 \text{ mmHg}}=30.6 \text{ L}\end{align*}

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 L.

### Summary

• The ideal gas law is used to calculate stoichiometry problems for gases.

### Review

1. Do we need gas conditions to be at STP to calculate stoichiometry problems?
2. Why do we want to determine the stoichiometry of these reactions?
3. What assumption are we making about the gases involved?

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