12 g of Mg (at. mass 24) will react completely with acid to give

$\begin{array}{l}\mathrm{Mg}+2\mathrm{HCl}\to {\mathrm{MgCl}}_2+{\mathrm{H}}_2\\ 1\mathrm{mole} \mathrm{of} \mathrm{Mg}(=24\mathrm{g})\to 1\mathrm{mole} \mathrm{of} {\mathrm{H}}_2\\ 12\mathrm{g} \mathrm{of} \mathrm{Mg}\to ? \\ {\mathrm{n}}_{{\mathrm{H}}_2}=\frac{12\mathrm{g}}{24\mathrm{g}}=\frac{1}{2}\mathrm{mole}\mathrm{of}{\mathrm{H}}_2\end{array}$