Testing

This lesson is a quick check-up on stoichiometry calculations — the maths that connects balanced chemical equations to real quantities like mass, volume, and moles. Before you test yourself, here's a fast recap of the four problem types you've met this month (Talbot, p.424).

Quick Recap: The Four Stoichiometry Problem Types

• Mass–mass: convert given mass → moles → moles of unknown (using equation ratio) → mass of unknown (Talbot, p.424)
• Mass–gas volume: convert given mass → moles → moles of unknown → volume using molar gas volume at STP (Talbot, p.425)
• Gas volume–gas volume: use mole ratio directly as volume ratio (no need to convert to moles first!) (Talbot, p.425)
• Concentration: uses n = c × V for reactions in solution

The golden rule for every type: always work through moles in the middle step, using the coefficients of the balanced equation as your mole ratio.

Worked Example Recap

Question: What volume of H₂ gas (at STP) forms when 1.64 g of aluminium reacts completely with excess hydrochloric acid?

Equation: 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

Steps:

1. n(Al) = m / M = 1.64 / 26.98 = 0.0608 mol
2. Mole ratio Al : H₂ = 2 : 3, so n(H₂) = 0.0608 × (3/2) = 0.0912 mol
3. V(H₂) = n × Vm = 0.0912 × 22.7 dm³ mol⁻¹ ≈ 2.07 dm³ (Talbot, p.425)

Notice how the calculation always follows: mass → moles → moles → mass/volume.

Key Takeaways

• All stoichiometry calculations pivot around converting to moles, since moles connect directly to the balanced equation's coefficients.
• Use n = m / M to convert mass to moles, and V = n × Vm (22.7 dm³ mol⁻¹ at STP) to convert moles to gas volume.
• For gas-to-gas reactions, you can skip straight to using the mole ratio as a volume ratio — no conversion needed.
• Always double-check your equation is balanced before starting any calculation — an unbalanced equation gives wrong ratios.
• Show your working step-by-step (mass→moles→moles→answer); partial credit and error-checking both depend on this.

Practice Set (5 Questions)

Q1. Calculate the amount (in mol) of CO₂ produced when 5.00 g of CaCO₃ decomposes completely: CaCO₃(s) → CaO(s) + CO₂(g) (M of CaCO₃ = 100.09 g mol⁻¹)

Q2. What volume of O₂ gas (at STP) is needed to completely combust 3.00 g of methane, CH₄? CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) (M of CH₄ = 16.05 g mol⁻¹)

Q3. In the reaction N₂(g) + 3H₂(g) → 2NH₃(g), what volume of NH₃ forms from 6.0 dm³ of H₂ at the same temperature and pressure?

Q4. Determine the mass of Mg needed to react completely with 0.500 mol of O₂: 2Mg(s) + O₂(g) → 2MgO(s) (M of Mg = 24.31 g mol⁻¹)

Q5. A student reacts 2.45 g of Zn with excess HCl. Calculate the volume of H₂ gas produced at STP. Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g) (M of Zn = 65.38 g mol⁻¹)

Answers

A1. n(CaCO₃) = 5.00 / 100.09 = 0.0500 mol → mole ratio 1:1 → n(CO₂) = 0.0500 mol

A2. n(CH₄) = 3.00 / 16.05 = 0.187 mol → mole ratio CH₄:O₂ = 1:2 → n(O₂) = 0.373 mol → V = 0.373 × 22.7 = 8.47 dm³

A3. Mole ratio H₂:NH₃ = 3:2 → V(NH₃) = 6.0 × (2/3) = 4.0 dm³

A4. Mole ratio O₂:Mg = 1:2 → n(Mg) = 0.500 × 2 = 1.00 mol → m = 1.00 × 24.31 = 24.3 g

A5. n(Zn) = 2.45 / 65.38 = 0.0375 mol → mole ratio Zn:H₂ = 1:1 → n(H₂) = 0.0375 mol → V = 0.0375 × 22.7 = 0.851 dm³