$Mg+2HCl\to MgC{{l}_{2}}+{{H}_{2}}$

0,5              $\to $           0,5    (mol)

${{n}_{Mg}}=\frac{{{m}_{Mg}}}{{{M}_{Mg}}}=\frac{12}{24}=0,5\left( mol \right)$

Theo phương trình ta có:

$\begin{align}  & {{n}_{Mg}}={{n}_{{{H}_{2}}}}=0,5\left( mol \right) \\ & {{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,5=11,2\left( l \right) \\\end{align}$

$\Rightarrow $Chọn đáp án C