Hướng dẫn: ${{n}_{HCl}}={{V}_{HCl}}.{{C}_{{{M}_{HCl}}}}=0,2.2=0,4\left( mol \right)$        

           $Mn{{O}_{2}}+4HCl\xrightarrow{{{t}^{o}}}MnC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O$ (1)

               0,4                    0,1                         (mol)

$C{{l}_{2}}+2NaOH\to NaCl+NaClO+{{H}_{2}}O$ (2)

0,1 0,2   (mol)

Theo phương trình (1) có: ${{n}_{C{{l}_{2}}}}=\frac{{{n}_{HCl}}}{4}=\frac{0,4}{4}=0,1\left( mol \right)$

Theo phương trình (2) có: ${{n}_{NaOH}}=2.{{n}_{C{{l}_{2}}}}=2.0,1=0,2\left( mol \right)$

$\Rightarrow {{V}_{NaOH}}=\frac{{{n}_{NaOH}}}{{{C}_{{{M}_{NaOH}}}}}=\frac{0,2}{5}=0,04\left( l \right)=40ml$

Chọn đáp án B