Hướng dẫn:

${{n}_{{{H}_{2}}}}=\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{1,12}{22,4}=0,05\left( mol \right)$

$2C{{H}_{3}}COOH+2Na\to 2C{{H}_{3}}COONa+{{H}_{2}}\uparrow $

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

Theo phương trình ta có: ${{n}_{C{{H}_{3}}COOH}}=2.{{n}_{{{H}_{2}}}}=2.0,05=0,1\left( mol \right)$

$\Rightarrow {{V}_{C{{H}_{3}}COOH}}=\frac{{{n}_{C{{H}_{3}}COOH}}}{{{C}_{M}}_{_{C{{H}_{3}}COOH}}}=\frac{0,1}{2}=0,05\left( l \right)=50\left( ml \right)$

$\Rightarrow $Chọn đáp án B