# Purity of calcium carbonate in limestone

$\mathrm{150~mL}$ of $\mathrm{0.1~N}$ $\ce{HCl}$ is required to react with $\mathrm{1~g}$ of a sample of limestone. Calculate percentage purity of calcium carbonate.

Reaction: $$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$

Number of milli equivalence of $\ce{HCl}$ used initially is $\mathrm{15}$

Number of milli moles of $\ce{HCl}$ used initially is $\mathrm{\frac{15}{2}}$

$\mathrm{2}$ moles of $\ce{HCl}$ reacts with $\mathrm{1}$ mole of $\ce{CaCO3}$.

So amount of calcium carbonate is $\mathrm{\frac{15}{4}\times 10^{-3}~mol}=\mathrm{\frac{3}{8}~g}$

But answer given is $\mathrm{75\%}$

We know that 2 moles of $\ce{HCl}$ are required to react completely with 1 mole of $\ce{CaCO3}$. We find out the number of moles of $\ce{HCl}$ in $\mathrm{150ml}$ $\mathrm{0.1M}$ $\ce{HCl}$ ($\mathrm{0.1N} \ce{HCl} = \mathrm{0.1Ml} \ce{HCl}$), which is $\mathrm{0.015}$ moles. Now the amount of $\ce{CaCO3}$ reacted is ($\mathrm{\frac{0.015}{2}}$) = $\mathrm{0.0075}$ moles.
Amount of $\ce{CaCO3}$ in gms = $$\mathrm{0.0075~moles\times 100~}\text{(mol.wt of CaCO3)} = \mathrm{0.75g}$$