Hướng dẫn:

${{n}_{{{H}_{2}}}}=\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{8,96}{22,4}=0,4\left( mol \right)$

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

0,4            $\leftarrow $              0,4                  (mol)

Theo phương trình ta có: ${{n}_{{{H}_{2}}}}={{n}_{Zn}}=0,4\left( mol \right)$

$\Rightarrow {{m}_{Zn}}={{n}_{Zn}}.{{M}_{Zn}}=0,4.65=26\left( g \right)$

 $\Rightarrow $ Chọn đáp án A