$\begin{align}  & {{n}_{Zn}}=\frac{{{m}_{Zn}}}{{{M}_{Zn}}}=\frac{9,75}{65}=0,15\left( mol \right) \\ & {{n}_{Mg}}=\frac{{{m}_{Mg}}}{{{M}_{Mg}}}=\frac{6}{24}=0,25\left( mol \right) \\ &  \\\end{align}$

$Zn+2HCl\to ZnC{{l}_{2}}+{{H}_{2}}$

0,15            $\to $               0,15     (mol)

$Fe+2HCl\to FeC{{l}_{2}}+{{H}_{2}}$

0,25         $\to $           0,25     (mol)

Theo phương trình ta có:

${{n}_{{{H}_{2}}}}={{n}_{Zn}}+{{n}_{Fe}}=0,15+0,25=0,4\left( mol \right)$

${{V}_{{{H}_{2}}}}=22,4.0,4=8,96\left( l \right)$

$\Rightarrow $Chọn đáp án D.