2025 H2 Chem P1 Q6 — Mole Solution

The answer is A (129 \(\text{cm}^3\)).
\(n(\text{Fe}) = \dfrac{0.422}{55.8} = 7.56 \times 10^{-3}\) mol
Link Fe back to \(\text{O}_2\) through the three equations:
• 2 mol Fe needs 3 mol CO (third equation).
• 3 mol CO comes from \(\tfrac{3}{2}\) mol \(\text{CO}_2\) (second equation: 1 \(\text{CO}_2\) gives 2 CO).
• \(\tfrac{3}{2}\) mol \(\text{CO}_2\) needs \(\tfrac{3}{2}\) mol \(\text{O}_2\) (first equation).
So \(2\text{Fe} \equiv 3\text{CO} \equiv \tfrac{3}{2}\text{CO}_2 \equiv \tfrac{3}{2}\text{O}_2\), i.e. \(n(\text{O}_2) = \tfrac{3}{4}\,n(\text{Fe})\).
\(n(\text{O}_2) = \tfrac{3}{4} \times 7.56 \times 10^{-3} = 5.67 \times 10^{-3}\) mol
\(V(\text{O}_2) = 5.67 \times 10^{-3} \times 22.7 = 0.129\ \text{dm}^3 = 129\ \text{cm}^3\)
• B is wrong: 229 \(\text{cm}^3\) comes from inverting the ratio, \(n(\text{O}_2) = \tfrac{4}{3} \times 7.56 \times 10^{-3} = 1.01 \times 10^{-2}\) mol, and \(1.01 \times 10^{-2} \times 22.7 = 0.229\ \text{dm}^3\).
• C is wrong: 258 \(\text{cm}^3\) uses \(n(\text{O}_2) = \tfrac{3}{2}\,n(\text{Fe})\), which ignores that each \(\text{CO}_2\) produces 2 CO in the second equation, so it doubles the oxygen needed.
• D is wrong: 386 \(\text{cm}^3\) is three times the correct volume (\(\tfrac{9}{4}\) mol \(\text{O}_2\) per mol Fe), which no combination of the three equations gives.