5g of a sample of magnesium carbonate on treatment with excess of dilute hydrochloric acid gave 1.12 L of CO₂ at STP. The percentage of magnesium carbonate in the mixture is

$$\begin{gathered} {\mathrm{MgCO}}_3+2\mathrm{HCl}\to {\mathrm{MgCl}}_2+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O} \\ 84\mathrm{g} 22.4\mathrm{L} \\ 5\mathrm{g} ? \end{gathered}$$

Volume of CO₂ liberated= $\frac{5}{84}\times 22.4=1.33\mathrm{L}$

The percentage of magnesium carbonate=$\frac{1.12}{1.33}\times 100=84\%$