Hướng dẫn:

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

Trong hỗn hợp A chỉ có axetic tác dụng với K.

$2C{{H}_{3}}COOH+2K\to 2C{{H}_{3}}COOK+{{H}_{2}}$

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

Theo phương trình ta có:

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

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

$\Rightarrow {{m}_{C{{H}_{4}}}}={{m}_{A}}-{{m}_{C{{H}_{3}}COOH}}=18,6-12=6,6\left( g \right)$

$%{{m}_{C{{H}_{4}}}}=\frac{{{m}_{C{{H}_{4}}}}}{{{m}_{A}}}.100=\frac{6,6}{18,6}.100=35,48\left( % \right)$

$\Rightarrow $Chọn đáp án A