Hướng dẫn:

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

$CO+{{O}_{2}}\xrightarrow{{{t}^{0}}}C{{O}_{2}}$       (1)

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

$C{{O}_{2}}+{{K}_{2}}O\to {{K}_{2}}C{{O}_{3}}$      (2)

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

Theo phương trình (1) có: ${{n}_{C{{O}_{2}}}}={{n}_{CO}}=0,15\left( mol \right)$

Theo phương trình (2) có: ${{n}_{{{K}_{2}}C{{O}_{3}}}}={{n}_{C{{O}_{2}}}}=0,15\left( mol \right)$

$\Rightarrow {{m}_{{{K}_{2}}C{{O}_{3}}}}={{n}_{{{K}_{2}}C{{O}_{3}}}}.{{M}_{{{K}_{2}}C{{O}_{3}}}}=0,15.138=20,7\left( g \right)$

$\Rightarrow $Chọn đáp án A