Hướng dẫn:

${{n}_{{{H}_{2}}}}=\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{3,36}{22,4}=0,15\left( mol \right)$

Gọi a,b lần lượt là số mol của etylic và axetic

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

    b                         $\leftarrow $                          $\frac{b}{2}$ (mol)

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

    a                     $\leftarrow $                      $\frac{a}{2}$ (mol)

Theo 2 phương trình ta có:

${{n}_{{{H}_{2}}}}=\frac{{{n}_{{{C}_{2}}{{H}_{5}}OH}}}{2}+\frac{{{n}_{C{{H}_{3}}COOH}}}{2}=\frac{a}{2}+\frac{b}{2}=0,15\left( mol \right)$

$\Rightarrow a+b=2.0,15=0,3\left( mol \right)$                                        (*)

$\Rightarrow {{m}_{A}}={{m}_{C{{H}_{3}}COOH}}+{{m}_{{{C}_{2}}{{H}_{5}}OH}}={{n}_{C{{H}_{3}}COOH}}.{{M}_{C{{H}_{3}}COOH}}+{{n}_{{{C}_{2}}{{H}_{5}}OH}}.{{M}_{{{C}_{2}}{{H}_{5}}OH}}$

$\Rightarrow 60.b+46.a=15,9\left( g \right)$                                               (**)

Từ (*) và (**) có: a= 0,15 (mol)

                             b= 0,15 (mol)

$\Rightarrow {{m}_{C{{H}_{3}}COOH}}={{n}_{C{{H}_{3}}COOH}}.{{M}_{C{{H}_{3}}COOH}}=0,15.60=9\left( g \right)$

$%{{m}_{C{{H}_{3}}COOH}}=\frac{{{m}_{C{{H}_{3}}COOH}}}{{{m}_{A}}}.100=\frac{9}{15}.100=60\left( % \right)$

$\Rightarrow $Chọn đáp án D