What mass of 95\% pure $\mathrm{CaCO}{}_3$ is required to neutralise 50 mL of 0.5 M $\mathrm{HCl}$ according to the reaction below? (answer to second decimal place) $\mathrm{CaCO}{}_3(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq})\to \mathrm{CaCl}{}_2(\mathrm{aq})+\mathrm{CO}{}_2(\mathrm{g})+2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})$

Solution & Explanation

\textbf{Answer:} (B) 1.32 g \textbf{Solution:} Moles of $\mathrm{HCl}$ $=0.5\times \frac{50}{1000}=0.025$ mol. From stoichiometry, 2 mol $\mathrm{HCl}$ $\equiv$ 1 mol $\mathrm{CaCO}{}_3$, so moles $\mathrm{CaCO}{}_3$ $=\frac{0.025}{2}=0.0125$ mol. Mass of pure $\mathrm{CaCO}{}_3$ $=0.0125\times 100=1.25$ g. For a 95\% pure sample: mass $=1.25\times \frac{100}{95}=1.32$ g.