Answer: B. Equation: \(\text{Pb(NO}_3\text{)}_2 + 2\text{NaCl} \rightarrow \text{PbCl}_2 + 2\text{NaNO}_3\). Sodium chloride is in excess, so the lead(II) nitrate is the limiting reagent.
Moles of \(\text{Pb(NO}_3\text{)}_2 = c \times V = 0.1 \times \dfrac{25}{1000} = 0.0025\) mol.
Mole ratio \(\text{Pb(NO}_3\text{)}_2 : \text{PbCl}_2 = 1 : 1\), so moles of \(\text{PbCl}_2 = 0.0025\) mol.
\(M_r(\text{PbCl}_2) = 207 + 2 \times 35.5 = 278\).
Mass = \(n \times M_r = 0.0025 \times 278 = 0.695\) g — option B. Option A (0.348 g) halves this (perhaps confusing moles of \(\text{PbCl}_2\) with moles of \(\text{Cl}^-\)). Options C and D are out by factors of 10 and 40, likely from arithmetic slips.
Every O Level Pure Chemistry TYS MCQ solution video, plus full Sec 3 / Sec 4 topic recordings.
Find out more