$N{{a}_{2}}O+2HCl\to 2NaCl+{{H}_{2}}O$

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

${{n}_{HCl}}={{C}_{{{M}_{HCl}}}}.{{V}_{HCl}}=0,5.0,4=0,2\left( mol \right)$

Theo phương trình có:   ${{n}_{N{{a}_{2}}O}}=\frac{{{n}_{HCl}}}{2}=\frac{0,2}{2}=0,1\left( mol \right)$

${{m}_{N{{a}_{2}}O}}={{n}_{N{{a}_{2}}O}}.{{M}_{N{{a}_{2}}O}}=0,1.62=6,2\left( g \right)$

$\Rightarrow $Chọn đáp án C