Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
N{a_2}O + {H_2}O \to 2NaOH\\
b)\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Na}} = 2{n_{{H_2}}} = 0,3mol\\
{m_{Na}} = 0,3 \times 23 = 6,9g\\
{m_{N{a_2}O}} = 13,1 - 6,9 = 6,2g\\
\% Na = \dfrac{{6,9}}{{13,1}} \times 100\% = 52,7\% \\
\% N{a_2}O = 100 - 52,7 = 47,3\%
\end{array}\)