Hướng dẫn:

${{n}_{{{C}_{2}}{{H}_{5}}OH}}={{V}_{{{C}_{2}}{{H}_{5}}OH}}.{{C}_{M}}_{_{{{C}_{2}}{{H}_{5}}OH}}=0,4.0,25=0,1\left( mol \right)$

  $2{{C}_{2}}{{H}_{5}}OH+2Na\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\uparrow $

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

Theo phương trình ta có: ${{n}_{{{H}_{2}}}}=\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{2}=\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{2}=\frac{0,1}{2}=0,05\left( mol \right)$

$\Rightarrow {{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,05=1,12\left( l \right)$

$\Rightarrow $Chọn đáp án A