- Chọn A.

- Ta có:  ${{n}_{{{H}_{2}}O}}=\frac{3,6}{18}=0,2\left( mol \right)$

             $4Na+{{O}_{2}}\to 2N{{a}_{2}}O\left( 1 \right)$

              $N{{a}_{2}}O+{{H}_{2}}O\to 2NaOH\left( 2 \right)$

              $0,2\left( mol \right)$       $\leftarrow $  $0,2\left( mol \right)$

Theo PTHH (1): ${{n}_{Na}}=2{{n}_{N{{a}_{2}}O}}=2\times 0,2=0,4\left( mol \right)$.

- ${{m}_{Na}}=0,4\times 23=9,2\left( gam \right)$