Hướng dẫn:

${{n}_{KOH}}={{V}_{KOH}}.{{C}_{{{M}_{KOH}}}}=0,5.0,2=0,1\left( mol \right)$

${{n}_{NaOH}}={{V}_{NaOH}}.{{C}_{{{M}_{NaOH}}}}=0,5.0,1=0,05\left( mol \right)$

$2KOH+C{{O}_{2}}\to {{K}_{2}}C{{O}_{3}}+{{H}_{2}}O$   (1)

 0,1 $\to $ 0,05                                      (mol)

$2NaOH+C{{O}_{2}}\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O$         (2)

0,05   $\to $0,025                                  (mol)

Theo phương trình (1) và (2) có:

$\sum{{{n}_{C{{O}_{2}}}}=\frac{{{n}_{KOH}}}{2}+\frac{{{n}_{NaOH}}}{2}=\frac{0,1}{2}+\frac{0,05}{2}=0,05}+0,025=0,075\left( mol \right)$

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

Chọn đáp án B