Hướng dẫn:

${{n}_{C{{l}_{2}}}}=\frac{{{V}_{C{{l}_{2}}}}}{22,4}=\frac{5,6}{22,4}=0,25\left( mol \right)$

$C{{l}_{2}}+{{H}_{2}}\to 2HCl$                                   (1)

0,25       0.5              (mol)

$HCl+NaOH\to NaCl+{{H}_{2}}O$             (2)

0,5 0,5                                   (mol)

Theo phương trình (1) có: ${{n}_{HCl}}=2.{{n}_{C{{l}_{2}}}}=2.0,25=0,5\left( mol \right)$

Theo phương trình (2) có: ${{n}_{NaOH}}={{n}_{HCl}}=0,5\left( mol \right)$

$\Rightarrow {{m}_{NaOH}}={{n}_{NaOH}}.{{M}_{NaOH}}=0,5.40=20\left( g \right)$

$\Rightarrow C%=\frac{{{m}_{NaOH}}}{{{m}_{\text{d}{{\text{d}}_{NaOH}}}}}.100=\frac{20}{200}.100=10\left( % \right)$

Chọn đáp án C