Answer: C. The reaction is \(\text{XO}_2 + 2\text{CO} \rightarrow \text{X} + 2\text{CO}_2\).
Moles of \(\text{CO}_2\) produced = \(\frac{862}{24000} = 0.0359\) mol (at room temperature and pressure, molar volume = 24 dm³).
From the 1:2 mole ratio between \(\text{XO}_2\) and \(\text{CO}_2\), moles of \(\text{XO}_2\) = \(0.0359 / 2 = 0.01795\) mol.
\(M_r(\text{XO}_2) = \frac{4.29}{0.01795} = 239\). Subtract 32 for the two oxygen atoms: \(M_r(\text{X}) = 239 - 32 = 207\).
This matches \({}^{207}_{82}\text{Pb}\) — lead(IV) oxide reduced by carbon monoxide gives lead.
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