5 g of a sample of Magnesium carbonate on treatment with excess of dilute hydrochloric acid gave 1.12 L of ${\mathrm{CO}}_2$ at STP.  The percentage of magnesium carbonate in the mixture is

$$\begin{gathered} \underset{84\mathrm{g}}{{\mathrm{MgCO}}_3}+2\mathrm{HCl}\to {\mathrm{MgCl}}_2+\underset{ 22.4\mathrm{L}}{{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2} \\ \mathrm{x}1.12\mathrm{L} \\ \mathrm{x}=\frac{84\times 1.12}{22.4}=4.2\mathrm{g} \end{gathered}$$

 $\mathrm{\%}$ of ${\mathrm{MgCO}}_3$ in the sample  $\frac{4.2}{5}\times 100=84$