$\begin{align}  & {{n}_{HCl}}={{V}_{HCl}}.{{C}_{{{M}_{HCl}}}}=0,5.0,2=0,1\left( mol \right) \\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}={{V}_{{{H}_{2}}S{{O}_{4}}}}.{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=0,5.0,3=0,15\left( mol \right) \\\end{align}$

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

         0,1   $\to $          0,05               (mol)

$Zn+{{H}_{2}}S{{O}_{4}}\to ZnS{{O}_{4}}+{{H}_{2}}$

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

Theo phương trình ta có:

 $\begin{align} & {{n}_{{{H}_{2}}}}=\frac{{{n}_{HCl}}}{2}+{{n}_{{{H}_{2}}S{{O}_{4}}}}=\frac{0,1}{2}+0,15=0,2\left( mol \right) \\ & \Rightarrow {{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,2=4,48\left( l \right) \\\end{align}$

$\Rightarrow $Chọn đáp án B.