Hướng dẫn:

${{n}_{C{{O}_{2}}}}=\frac{{{V}_{C{{O}_{2}}}}}{22,4}=\frac{2,24}{22,4}=0,1\left( mol \right)$

$2C{{H}_{3}}COOH+N{{a}_{2}}C{{O}_{3}}\to 2C{{H}_{3}}COONa+{{H}_{2}}O+C{{O}_{2}}\uparrow $

  0,2                                $\leftarrow $                                      0,1         (mol)

Theo phương trình ta có:

${{n}_{C{{H}_{3}}COOH}}=2.{{n}_{C{{O}_{2}}}}=2.0,1=0,2\left( mol \right)$

$\Rightarrow {{m}_{C{{H}_{3}}COOH}}={{n}_{C{{H}_{3}}COOH}}.{{M}_{C{{H}_{3}}COOH}}=0,2.60=12\left( g \right)$

$\Rightarrow $Chọn đáp án B