Example

This lesson applies the four-step stoichiometry method to worked examples: mass to mass, the mass of a reactant needed, a gas volume, a number of particles and an industrial-scale calculation.

Every stoichiometric calculation follows the same path: write the balanced equation, convert the given quantity into moles, use the mole ratio, then convert into the unit the question asks for. This lesson works through six examples that use the three key conversions (n = m ÷ M for masses, V = n × 22.7 dm³ mol⁻¹ for gases at STP, and N = n × L for particles) and shows how to set out an answer clearly, with units and a sensible number of significant figures.

Stoichiometric Calculations: Worked Examples

Every calculation in this lesson follows the same four steps:

1. Write the balanced equation.
2. Convert the quantity you are given into moles.
3. Use the mole ratio (the coefficients) to find the moles of the substance you want.
4. Convert those moles into the unit the question asks for.

The conversions you need:

• Mass: n = m ÷ M, where m is the mass in grams and M is the molar mass in g mol⁻¹.
• Gas volume at standard temperature and pressure (STP: 273 K and 100 kPa): V = n × 22.7 dm³ mol⁻¹ for an ideal gas.
• Particles: N = n × L, where the Avogadro constant L = 6.02 × 10²³ mol⁻¹.

Relative atomic masses used: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Mg = 24.3, Ca = 40.1, Fe = 55.8.

Worked problem 1: mass of product from mass of reactant What mass of carbon dioxide forms when 8.0 g of methane burns completely?

1. CH₄ + 2O₂ → CO₂ + 2H₂O
2. M(CH₄) = 12.0 + (4 × 1.0) = 16.0 g mol⁻¹, so n(CH₄) = 8.0 ÷ 16.0 = 0.50 mol
3. CH₄ : CO₂ = 1 : 1, so n(CO₂) = 0.50 mol
4. M(CO₂) = 12.0 + (2 × 16.0) = 44.0 g mol⁻¹, so m(CO₂) = 0.50 × 44.0 = 22 g

Worked problem 2: mass of a reactant needed What mass of magnesium is needed to make 10.0 g of magnesium oxide?

1. 2Mg + O₂ → 2MgO
2. M(MgO) = 24.3 + 16.0 = 40.3 g mol⁻¹, so n(MgO) = 10.0 ÷ 40.3 = 0.248 mol
3. Mg : MgO = 2 : 2 = 1 : 1, so n(Mg) = 0.248 mol
4. m(Mg) = 0.248 × 24.3 = 6.03 g

Worked problem 3: volume of a gas Strongly heating calcium carbonate (limestone) produces calcium oxide and carbon dioxide. What volume of carbon dioxide, measured at STP, comes from 25.0 g of calcium carbonate?

1. CaCO₃ → CaO + CO₂
2. M(CaCO₃) = 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹, so n(CaCO₃) = 25.0 ÷ 100.1 = 0.2498 mol
3. CaCO₃ : CO₂ = 1 : 1, so n(CO₂) = 0.2498 mol
4. V(CO₂) = 0.2498 × 22.7 = 5.67 dm³

Worked problem 4: number of particles How many water molecules form when 0.25 mol of hydrogen gas burns in oxygen?

1. 2H₂ + O₂ → 2H₂O
2. The amount is already in moles: n(H₂) = 0.25 mol
3. H₂ : H₂O = 2 : 2 = 1 : 1, so n(H₂O) = 0.25 mol
4. N = 0.25 × 6.02 × 10²³ = 1.5 × 10²³ molecules

Worked problem 5: a mole ratio that is not 1 : 1 What mass of ammonia can be made from 6.0 g of hydrogen gas if nitrogen is in excess?

1. N₂ + 3H₂ → 2NH₃
2. M(H₂) = 2.0 g mol⁻¹, so n(H₂) = 6.0 ÷ 2.0 = 3.0 mol
3. H₂ : NH₃ = 3 : 2, so n(NH₃) = 3.0 × 2 ÷ 3 = 2.0 mol
4. M(NH₃) = 14.0 + (3 × 1.0) = 17.0 g mol⁻¹, so m(NH₃) = 2.0 × 17.0 = 34 g

Worked problem 6: scaling up to industry What mass of iron can a blast furnace obtain from 1.00 kg of iron(III) oxide?

1. Fe₂O₃ + 3CO → 2Fe + 3CO₂
2. M(Fe₂O₃) = (2 × 55.8) + (3 × 16.0) = 159.6 g mol⁻¹, so n(Fe₂O₃) = 1000 ÷ 159.6 = 6.266 mol
3. Fe₂O₃ : Fe = 1 : 2, so n(Fe) = 2 × 6.266 = 12.53 mol
4. m(Fe) = 12.53 × 55.8 = 699 g, or about 0.699 kg

A quick check: iron makes up 111.6 g of every 159.6 g of the oxide, which is about 70%, so roughly 0.7 kg of iron from 1 kg of oxide is sensible.

Tips for clear answers

• Always start from the balanced equation; a wrong ratio spoils every step after it.
• Keep units with every number, carry an extra digit through the working, and round only at the end.
• Give the final answer to the same number of significant figures as the data, usually two or three.
• Ask whether the answer is reasonable before moving on.