Hướng dẫn:

${{m}_{C{{H}_{3}}COOH}}=\frac{{{m}_{\text{d}{{\text{d}}_{C{{H}_{3}}COOH}}}}.C%}{100}=\frac{500.1,8}{100}=9\left( g \right)$

${{n}_{C{{H}_{3}}COOH}}=\frac{{{m}_{C{{H}_{3}}COOH}}}{{{M}_{C{{H}_{3}}COOH}}}=\frac{9}{60}=0,15\left( mol \right)$

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

   0,15                       $\to $                            0,075  (mol)

Theo phương trình ta có: ${{n}_{{{H}_{2}}}}=\frac{{{n}_{C{{H}_{3}}COOH}}}{2}=\frac{0,15}{2}=0,075\left( mol \right)$

${{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,075=1,68\left( l \right)$

$\Rightarrow $Chọn đáp án B