Core Concept

Gas stoichiometry is the study of the relationships between volumes, moles, and masses of gaseous reactants and products in a chemical reaction. Using gas laws and stoichiometric principles, you can predict the amounts of gases involved in reactions, whether they are measured in moles, liters, or grams.

Key Concepts

1. Ideal Gas Law:

   • The ideal gas law relates pressure, volume, temperature, and moles of a gas: PV=nRTPV = nRTPV=nRT

   • P = Pressure (usually in atm), V = Volume (L), n = Moles of gas, R = Ideal gas constant (0.0821 L⋅atm K−1 mol−10.0821 \, \text{L} \cdot \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1}0.0821L⋅atmK−1mol−1), T = Temperature (K).

2. Molar Volume of a Gas at STP:

   • At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 L.

   • STP Conditions: 0∘C0^\circ\text{C}0∘C (273.15 K) and 1 atm pressure.

3. Stoichiometry and Mole Ratios:

   • Just like in standard stoichiometry, the coefficients in a balanced chemical equation represent the mole ratios of the reactants and products.

   • These mole ratios allow you to relate moles of one substance to moles of another.

4. Volume Ratios:

   • In reactions where all reactants and products are gases at the same temperature and pressure, you can use the mole ratio as a volume ratio.

   • For example, in the reaction 2H2+O2→2H2O\text{2H}_2 + \text{O}_2 \rightarrow \text{2H}_2\text{O}2H2​+O2​→2H2​O, 2 liters of H2\text{H}_2H2​ react with 1 liter of O2\text{O}_2O2​ to produce 2 liters of H2O\text{H}_2\text{O}H2​O vapor.

Steps for Solving Gas Stoichiometry Problems

1. Write and Balance the Chemical Equation:

   • Start with a balanced chemical equation to find the mole ratios.

2. Convert Known Quantities to Moles:

   • If you’re given grams of a gas, convert to moles using molar mass.

   • If you’re given volume (at STP), convert to moles using the molar volume (22.4 L/mol).

   • For other conditions, use the ideal gas law to find moles.

3. Use Mole Ratios to Find Unknown Quantities:

   • Use the mole ratio from the balanced equation to relate the known substance to the unknown substance.

4. Convert Moles of Desired Gas to Required Units:

   • Convert moles to volume at STP (using 22.4 L/mol) or use the ideal gas law if the gas is not at STP.

   • If mass is required, multiply moles by molar mass.

Example Problem: Gas Stoichiometry at STP

Problem: How many liters of CO2\text{CO}_2CO2​ are produced when 5.0 g of C2H5OH\text{C}_2\text{H}_5\text{OH}C2​H5​OH (ethanol) combusts completely at STP?

C2H5OH+3O2→2CO2+3H2O\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}C2​H5​OH+3O2​→2CO2​+3H2​O

Solution:

1. Write and Balance the Equation:

   • The equation is already balanced.

2. Convert Given Mass to Moles:

   • Molar mass of C2H5OH=46.07 g/mol\text{C}_2\text{H}_5\text{OH} = 46.07 \, \text{g/mol}C2​H5​OH=46.07g/mol: Moles of C2H5OH=5.0 g46.07 g/mol=0.108 mol\text{Moles of } \text{C}_2\text{H}_5\text{OH} = \frac{5.0 \, \text{g}}{46.07 \, \text{g/mol}} = 0.108 \, \text{mol}Moles of C2​H5​OH=46.07g/mol5.0g​=0.108mol

3. Use Mole Ratio to Find Moles of CO2\text{CO}_2CO2​:

   • From the balanced equation, 1 mol C2H5OH1 \, \text{mol C}_2\text{H}_5\text{OH}1mol C2​H5​OH produces 2 mol CO22 \, \text{mol CO}_22mol CO2​. 0.108 mol C2H5OH×2 mol CO21 mol C2H5OH=0.216 mol CO20.108 \, \text{mol C}_2\text{H}_5\text{OH} \times \frac{2 \, \text{mol CO}_2}{1 \, \text{mol C}_2\text{H}_5\text{OH}} = 0.216 \, \text{mol CO}_20.108mol C2​H5​OH×1mol C2​H5​OH2mol CO2​​=0.216mol CO2​

4. Convert Moles of CO2\text{CO}_2CO2​ to Volume at STP:

   • Since we are at STP, use the molar volume (22.4 L/mol) to convert to volume. 0.216 mol CO2×22.4 L/mol=4.84 L CO20.216 \, \text{mol CO}_2 \times 22.4 \, \text{L/mol} = 4.84 \, \text{L CO}_20.216mol CO2​×22.4L/mol=4.84L CO2​

Answer: 4.84 liters of CO2\text{CO}_2CO2​ are produced at STP.

Example Problem: Gas Stoichiometry Using Ideal Gas Law

Problem: How many grams of O2\text{O}_2O2​ are needed to produce 15.0 L of NO2\text{NO}_2NO2​ at 2.00 atm and 298 K in the following reaction?

2NO+O2→2NO22\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_22NO+O2​→2NO2​

Solution:

1. Write and Balance the Equation:

   • The equation is balanced as written.

2. Convert Volume of NO2\text{NO}_2NO2​ to Moles Using the Ideal Gas Law:

   • Rearrange the ideal gas law to solve for nnn: n=PVRTn = \frac{PV}{RT}n=RTPV​.

   • Given: P=2.00 atmP = 2.00 \, \text{atm}P=2.00atm, V=15.0 LV = 15.0 \, \text{L}V=15.0L, T=298 KT = 298 \, \text{K}T=298K, R=0.0821 L⋅atm K−1 mol−1R = 0.0821 \, \text{L} \cdot \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1}R=0.0821L⋅atmK−1mol−1. n=(2.00 atm)(15.0 L)(0.0821 L⋅atm K−1 mol−1)(298 K)=1.23 mol NO2n = \frac{(2.00 \, \text{atm})(15.0 \, \text{L})}{(0.0821 \, \text{L} \cdot \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1})(298 \, \text{K})} = 1.23 \, \text{mol NO}_2n=(0.0821L⋅atmK−1mol−1)(298K)(2.00atm)(15.0L)​=1.23mol NO2​

3. Use Mole Ratio to Find Moles of O2\text{O}_2O2​:

   • From the balanced equation, 2 mol NO22 \, \text{mol NO}_22mol NO2​ requires 1 mol O21 \, \text{mol O}_21mol O2​. 1.23 mol NO2×1 mol O22 mol NO2=0.615 mol O21.23 \, \text{mol NO}_2 \times \frac{1 \, \text{mol O}_2}{2 \, \text{mol NO}_2} = 0.615 \, \text{mol O}_21.23mol NO2​×2mol NO2​1mol O2​​=0.615mol O2​

4. Convert Moles of O2\text{O}_2O2​ to Grams:

   • Molar mass of O2=32.00 g/mol\text{O}_2 = 32.00 \, \text{g/mol}O2​=32.00g/mol: 0.615 mol O2×32.00 g/mol=19.68 g O20.615 \, \text{mol O}_2 \times 32.00 \, \text{g/mol} = 19.68 \, \text{g O}_20.615mol O2​×32.00g/mol=19.68g O2​

Answer: 19.68 grams of O2\text{O}_2O2​ are needed.

Key Tips for Gas Stoichiometry

• Use Ideal Gas Law for Non-STP Conditions: When conditions differ from STP, use PV=nRTPV = nRTPV=nRT to find moles or volume.

• Use Volume Ratios at STP: At STP, you can use the mole ratio as a volume ratio for gases.

• Convert Units Carefully: Always check if you need to convert between grams, moles, or liters.

• Temperature in Kelvin: Always use Kelvin in gas law calculations.