2019 O Level Chem P1 Q12 — Stoichiometry / Relative Atomic Mass Solution

Answer: C (¹¹⁹₅₀Sn, tin). The oxide XO₂ (4.00 g) is reduced to metal X (3.15 g), so the mass of oxygen removed is \(4.00 - 3.15 = 0.85\,\text{g}\).

Moles of oxygen atoms removed = \(0.85/16.0 = 0.0531\) mol. In XO₂ the ratio of X to O is 1:2, so moles of X = \(0.0531/2 = 0.0266\) mol.

The relative atomic mass of X is therefore \(3.15/0.0266 \approx 119\). The element with \(A_r \approx 119\) is tin, Sn (\(A_r = 118.7\)), so X is tin, option C.

Checking the distractors: titanium (48) and manganese (55) are far too light, while lead (207) is far too heavy — 0.0266 mol of lead would weigh about 5.5 g, not 3.15 g. Only tin gives a relative atomic mass consistent with the reacting masses.