Hướng dẫn:

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

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

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

Theo phương trình ta có:

${{n}_{C{{O}_{2}}}}=\frac{{{n}_{C{{H}_{3}}COOH}}}{2}=\frac{0,15}{2}=0,075\left( mol \right)$

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

$\Rightarrow $Chọn đáp án B