Gas Stoichiometry in Reactions

Gas stoichiometry is the use of balanced chemical equations together with the gas laws to calculate the volume, pressure or amount of gases consumed or produced in a reaction. In brief: convert the quantity you know into moles, use the mole ratio from the balanced equation, then convert the moles of the wanted gas into a volume with V = nRT/P (or 22.4 L/mol at STP).

Many reactions release or consume gases: metals dissolving in acid, the decomposition of carbonates, combustion and the inflation of airbags. Because gas volumes are easy to measure, they are a convenient way to follow a reaction quantitatively. This article gives a general method, four worked examples and a summary of common errors.

What is the method for gas stoichiometry?

  1. Write and balance the chemical equation.
  2. Convert the given quantity (mass, volume or pressure) to moles.
  3. Use the coefficients of the balanced equation to find the moles of the substance you want.
  4. Convert these moles to the requested unit, using the molar mass for a mass or the ideal gas law for a gas volume.
  5. Check that the answer has appropriate units and is reasonable.

Which conversions are needed?

Conversions between quantities in a reaction
From To moles Tool
Mass of solid or liquidn = m / MMolar mass
Gas at STPn = V / 22.414 L/molMolar volume
Gas at any conditionsn = PV / RTIdeal gas law
Solutionn = c × VMolarity
Other substance in the reactionnB = nA × (coefficient B / coefficient A)Mole ratio

Worked example: a solid decomposes to give a gas

Potassium chlorate decomposes on heating: 2KClO3(s) → 2KCl(s) + 3O2(g). What volume of oxygen at 25 °C and 1.00 atm is produced from 5.00 g of KClO3?

5.00 g KClO3 ÷122.55 0.0408 mol KClO3 ×3/2 0.0612 mol O2 RT/P 1.50 L O2
  1. The molar mass of KClO3 is 39.10 + 35.45 + 3 × 16.00 = 122.55 g/mol.
  2. n(KClO3) = 5.00 / 122.55 = 0.04080 mol.
  3. n(O2) = 0.04080 × 3/2 = 0.06120 mol.
  4. V = nRT / P = 0.06120 × 0.08206 × 298.15 / 1.00 = 1.50 L.

Worked example: gas volumes in a gas reaction

For the combustion 2CO(g) + O2(g) → 2CO2(g), how much oxygen is needed to burn 10.0 L of carbon monoxide, and how much carbon dioxide forms? Both gases are measured at the same temperature and pressure.

When all gases are at the same temperature and pressure, volume ratios are equal to mole ratios, so no gas law calculation is needed. The ratio is 2 : 1 : 2, so 10.0 L of CO requires 5.0 L of O2 and forms 10.0 L of CO2.

2CO 10.0 L + O2 5.0 L → 2CO2 10.0 L

Worked example: a gas collected over water

Magnesium reacts with hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g). A 1.50 g piece of magnesium is dissolved completely and the hydrogen is collected over water at 25 °C, where the total pressure is 745 mmHg. What volume of gas is collected? The vapor pressure of water at 25 °C is 23.8 mmHg.

1.50 g Mg ÷24.305 0.0617 mol Mg ×1 0.0617 mol H2 RT/P 1.59 L H2
  1. n(Mg) = 1.50 / 24.305 = 0.06171 mol, so n(H2) = 0.06171 mol.
  2. P(H2) = 745 − 23.8 = 721.2 mmHg = 721.2 / 760 = 0.9489 atm.
  3. V = nRT / P = 0.06171 × 0.08206 × 298.15 / 0.9489 = 1.51 / 0.9489 = 1.59 L.

Worked example: an airbag

Older car airbags use the decomposition of sodium azide: 2NaN3(s) → 2Na(s) + 3N2(g). How many liters of nitrogen at 300 K and 1.00 atm are released by 65.0 g of NaN3?

65.0 g NaN3 ÷65.02 1.00 mol NaN3 ×3/2 1.50 mol N2 RT/P 36.9 L N2
  1. M(NaN3) = 22.99 + 3 × 14.01 = 65.02 g/mol, so n = 65.0 / 65.02 = 1.00 mol.
  2. n(N2) = 1.00 × 3/2 = 1.50 mol.
  3. V = 1.50 × 0.08206 × 300 / 1.00 = 36.9 L.

This is roughly the volume of a driver’s airbag. Modern designs use other propellants, since sodium metal is a hazardous by-product that must be neutralized by additional reagents.

Why did gas volumes matter historically?

In 1808 Joseph Louis Gay-Lussac found that gases combine in small whole-number volume ratios, such as 2 volumes of hydrogen with 1 volume of oxygen. This observation, called the law of combining volumes, was difficult to explain until Avogadro’s hypothesis of 1811 linked equal volumes to equal numbers of molecules. These ideas let chemists deduce molecular formulas such as H2O and NH3 from measurements of gas volumes, and so they were instrumental in the development of reliable atomic masses.

Which mistakes are common?