Hướng dẫn:

$S{{O}_{2}}+2NaOH\to N{{a}_{2}}S{{O}_{3}}+{{H}_{2}}O$

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

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

0,15$\to $0,3                                       (mol)

$\begin{align}  & {{n}_{S{{O}_{2}}}}=\frac{{{V}_{S{{O}_{2}}}}}{22,4}=\frac{2,24}{22,4}=0,1\left( mol \right) \\ & {{n}_{C{{O}_{2}}}}=\frac{{{V}_{C{{O}_{2}}}}}{22,4}=\frac{3,36}{22,4}=0,15\left( mol \right) \\\end{align}$

Theo phương trình ta có: ${{n}_{NaOH}}=2.{{n}_{S{{O}_{2}}}}+2.{{n}_{C{{O}_{2}}}}=2.0,1+2.0,15=0,5\left( mol \right)$

${{V}_{NaOH}}=\frac{{{n}_{NaOH}}}{{{C}_{{{M}_{NaOH}}}}}=\frac{0,5}{0,2}=2,5(l)$

$\Rightarrow $Chọn đáp án B