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

No. of moles of $MgC{O}_3$ = $\frac{\mathrm{Given} \mathrm{volume}}{\mathrm{gram} \mathrm{molar} \mathrm{volume}}=\frac{1.12}{22.4}=0.5$

Weight of pure $MgC{O}_3$ =$84 \mathrm{x} 0.5= 4.2\mathrm{g}$

Percentage of $MgC{O}_3$ in sample= $4.2/ 5 x 100 = 84$

Hence, the correct option is C.