What mass of 95 % pure CaCO 3 will be required to neutralise 50 mL of 0 . 5 M HCl solution according to the following reaction? CaCO 3 s + 2 HCl aq → CaCl 2 aq + CO 2 g + 2 H 2 O l [Calculate upto second place of decimal point]

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Aicharya · Accepted Answer

$\underset{(s)}{\mathrm{CaCO}_3}+\underset{50 \mathrm{~mL}, 0.5 \mathrm{M}}{2 \mathrm{HCl}(\mathrm{aq})} \longrightarrow \underset{(\mathrm{aq})}{\mathrm{CaCl}_2}+\underset{(g)}{\mathrm{CO}_2}+\underset{(l)}{2 \mathrm{H}_2 \mathrm{O}}$ the number of moles of HCl taken $=0.5 \times 0.05$ $=0.025$ moles As we can see that from the above balanced equation is one mole of $\mathrm{CaCO}_3(s)$ requires 2 moles of $\mathrm{HCl}(\mathrm{aq})$ therefore, for 0.025 moles of $\mathrm{HCl}(\mathrm{aq}), 0.0125$ moles of $\mathrm{CaCO}_3(s)$ will be required. Mass of 0.0125 moles of $\mathrm{CaCO}_3(s)=0.0125 \times$ molar mass of $\mathrm{CaCO}_3$ $\begin{aligned} & =0.0125 \times 100 \ & =1.25 \mathrm{~g} \end{aligned}$ But purity of $\mathrm{CaCO}_3(s)$ is $95 \%$. Therefore, the actual amount of $\mathrm{CaCO}_3(s)$ required is Percentage purity $=\frac{\text { weight of pure substance }}{\text { weight of impure sample }} \times 100$ $95=\frac{1.25}{\text { weight of impure sample }} \times 100$ weight of impure sample $=\frac{1.25 \times 100}{95}$ $=1.32 \mathrm{~g}$ Hence. option 1 is correct.

What mass of 95 % pure CaCO 3 will be required to neutralise 50 mL of 0 . 5 M HCl solution according to the…

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