Gas Stoichiometry at STP and Non-STP

Gas stoichiometry applies mole ratios to reactions involving gases, using the volume of gas to find moles: at STP use the molar volume (22.4 L/mol), and under other conditions use the ideal gas law, n = PV ÷ RT. In brief: in N₂ + 3H₂ → 2NH₃, 15.0 L of hydrogen reacts with 5.00 L of nitrogen to give 10.0 L of ammonia at the same temperature and pressure, because the volume ratio of gases equals the mole ratio.

Many reactions produce or consume gases, such as combustion, the decomposition of carbonates and reactions of metals with acids. The volume of a gas depends on the temperature and the pressure, so gas stoichiometry needs to include these conditions. This article explains the two methods, the shortcut of volume ratios, how to correct for the vapor pressure of water when gas is collected over water, shows a table of gas laws, works through examples under various conditions and gives the history of the gas laws.

How do you do gas stoichiometry at STP?

At STP (0 °C and 1 atm), one mole of an ideal gas occupies 22.4 L. This value can be used as a conversion factor: V = n × 22.4 L/mol, or n = V ÷ 22.4 L/mol. The calculation follows the usual three steps: volume (or mass) of the given substance → moles → moles of the wanted substance (mole ratio) → volume (or mass) of the wanted substance.

What is the volume ratio shortcut?

According to Avogadro’s law, at the same temperature and pressure, the volumes of gases are proportional to the numbers of moles. So for a reaction in which all reactants and products are gases at the same temperature and pressure, the ratio of volumes equals the ratio of coefficients. This is Gay-Lussac’s law of combining volumes. For N₂ + 3H₂ → 2NH₃, 1 L of N₂ reacts with 3 L of H₂ to give 2 L of NH₃. No conversion to moles is needed. The shortcut works only for gases at the same temperature and pressure, and not for solids or liquids.

How do you handle non-standard conditions?

Use the ideal gas law PV = nRT. The units must be consistent: with P in atm and V in L, R = 0.082057 L·atm/(mol·K). With P in kPa and V in L, R = 8.314 L·kPa/(mol·K). T must be in kelvin.

1. Find the moles of the given substance (from mass, or from PV = nRT if it is a gas).
2. Use the mole ratio to find the moles of the gas wanted.
3. Use V = nRT ÷ P to find its volume at the stated conditions.

What if the gas is collected over water?

A gas that is collected by displacement of water is mixed with water vapor. The total pressure is the sum of the pressure of the gas and the vapor pressure of water at that temperature (Dalton’s law of partial pressures): P_total = P_gas + P_H2O. To find the moles of the gas, use P_gas = P_total − P_H2O.

Gas laws summary

Worked examples

1. Volume ratio. How much oxygen is needed to burn 4.00 L of propane gas, and how much CO₂ forms, C₃H₈ + 5O₂ → 3CO₂ + 4H₂O (same T and P, water as vapor)? O₂: 4.00 × 5 = 20.0 L. CO₂: 4.00 × 3 = 12.0 L. Water vapor: 4.00 × 4 = 16.0 L.
2. At STP. What volume of hydrogen forms at STP when 3.27 g of zinc reacts with excess HCl? n(Zn) = 3.27 ÷ 65.38 = 0.0500 mol. n(H₂) = 0.0500 mol. V = 0.0500 × 22.4 = 1.12 L.
3. Gas volume to mass at STP. What mass of CaCO₃ is needed to produce 5.60 L of CO₂ at STP? n(CO₂) = 5.60 ÷ 22.4 = 0.250 mol. n(CaCO₃) = 0.250 mol. m = 0.250 × 100.09 = 25.0 g.
4. Non-standard conditions. What volume of H₂ forms at 25 °C and 0.980 atm when 2.43 g of Mg reacts with excess acid? n(Mg) = 2.43 ÷ 24.305 = 0.1000 mol. n(H₂) = 0.1000 mol. V = nRT ÷ P = 0.1000 × 0.082057 × 298.15 ÷ 0.980 = 2.50 L.
5. Gas to moles under other conditions. A reaction yields 3.60 L of CO₂ at 30 °C and 101.3 kPa. n = PV ÷ RT = 101.3 × 3.60 ÷ (8.314 × 303.15) = 0.1447 mol, which corresponds to 6.37 g of CO₂.
6. Decomposition of potassium chlorate. What volume of O₂ at 20 °C and 101.3 kPa forms from 12.25 g of KClO₃, 2KClO₃ → 2KCl + 3O₂? n(KClO₃) = 12.25 ÷ 122.55 = 0.09996 mol. n(O₂) = 0.09996 × 1.5 = 0.1499 mol. V = 0.1499 × 8.314 × 293.15 ÷ 101.3 = 3.61 L.
7. Gas collected over water. Oxygen from the decomposition of KClO₃ is collected over water at 25 °C. The volume is 85.0 mL and the total pressure is 101.0 kPa. P(O₂) = 101.0 − 3.17 = 97.8 kPa. n = PV ÷ RT = 97.8 × 0.0850 ÷ (8.314 × 298.15) = 3.35 × 10⁻³ mol, which is 0.107 g of O₂. The mass of KClO₃ that decomposed is 3.35 × 10⁻³ × (2 ÷ 3) × 122.55 = 0.274 g.

Where is gas stoichiometry used?

• Airbags: the quantity of sodium azide needed to produce the volume of nitrogen that fills the bag.
• Combustion engineering: air-fuel ratios and exhaust volume.
• Breathing and medicine: the oxygen delivered by tanks and the carbon dioxide exhaled.
• Industrial chemistry: the volumes of gases in the production of ammonia, sulfuric acid and hydrogen.
• Environmental science: the volume of greenhouse gases released by burning fuels.

How were the gas laws discovered?

Robert Boyle established the relation between pressure and volume in 1662. Jacques Charles studied the effect of temperature in the 1780s, and Joseph Louis Gay-Lussac published his results in 1802 and his law of combining volumes in 1808. Amedeo Avogadro explained it in 1811. In 1834 Émile Clapeyron combined the laws in the ideal gas law. John Dalton’s law of partial pressures dates from 1801.

Which mistakes are common?

Mistake 1: Using 22.4 L/mol at any temperature

This value is only for 0 °C and 1 atm. For other conditions, use PV = nRT.

Mistake 2: Using Celsius in the ideal gas law

Convert to kelvin.

Mistake 3: Forgetting the water vapor pressure

When a gas is collected over water, subtract the vapor pressure of water.

Mistake 4: Applying the volume ratio to solids and liquids

The volume ratio works only for gases at the same temperature and pressure.