12 g of Mg (at. mass 24) on reacting completely with acid gives hydrogen gas, the volume of which at STP would be

$\begin{array}{l}\mathrm{Mg}+2\mathrm{HCl}\to {\mathrm{MgCl}}_2+{\mathrm{H}}_2\\ \because 24\mathrm{g}\mathrm{Mg}\mathrm{evolves}22.4\mathrm{L}{\mathrm{H}}_2\mathrm{at}\mathrm{STP}\\ \therefore \mathrm{volume} \mathrm{of} {\mathrm{H}}_2 \mathrm{evolved} \mathrm{by}12\mathrm{g}\text{\hspace{0.17em}of }\mathrm{Mg}\mathrm{at}\mathrm{STP}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{22.4}{24}\times 12=11.2\mathrm{L}\mathrm{at}\mathrm{STP}.\end{array}$