A mixture of Mg metal and $\mathrm{MgO}$ weighs 10 grams.  On treatment with excess of dilute $\mathrm{HCl}$, 2.24 L of hydrogen is produced at S.T.P.  The percentage of $\mathrm{MgO}$ present in the initial mixture is

 Of the two [$\mathrm{Mg}$ and $\mathrm{MgO}$] only $\mathrm{Mg}$ can liberate ${\mathrm{H}}_2$ with$\mathrm{HCl}$

$$\begin{gathered} \mathrm{From} \mathrm{eqution}: \\ 24\mathrm{g} \mathrm{of} \mathrm{Mg} \to 22.4\mathrm{l} \mathrm{of} {\mathrm{H}}_2 \mathrm{at} \mathrm{STP} \end{gathered}$$

amount of Mg required to produce 2.24l of ${\mathrm{H}}_2$ = 2.4g

$\mathrm{Mg} +$ $\mathrm{MgO}$ = 10 g

    $\mathrm{MgO}$ =10-2.4= 7.6 g     i.e., 76%