2021 O Level Chem P1 Q12 — Stoichiometry / Limiting Reagent Solution

Answer: A (120 cm³). The equation is Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.

\(n(\text{Na}_2\text{CO}_3) = 2.65 / 106 = 0.0250\,\text{mol}\); \(n(\text{HCl}) = 0.10 \times \tfrac{100}{1000} = 0.0100\,\text{mol}\).

Reacting all the carbonate would need \(2 \times 0.0250 = 0.0500\,\text{mol}\) of HCl, but only 0.0100 mol is present, so HCl is the limiting reagent. \(n(\text{CO}_2) = n(\text{HCl})/2 = 0.00500\,\text{mol}\), so \(V = 0.00500 \times 24\,000 = 120\,\text{cm}^3\) at room temperature and pressure.

The classic trap is to assume the carbonate is limiting, giving \(0.025 \times 24\,000 = 600\,\text{cm}^3\) (option C). Always check which reagent runs out first.