Giúp e với ạ e cám ơn ạ e cần gấp ạ

Em tham khảo nha:

\(\begin{array}{l}
1\,C\\
2\,C\\
3\,\\
2NaCl + 2{H_2}O \to 2NaOH + C{l_2} + {H_2}\\
4)\\
a)\\
2NaCl \to 2Na + C{l_2}\\
{n_{C{l_2}}} = \dfrac{{17,92}}{{22,4}} = 0,8\,mol\\
{n_{Na}} = 2{n_{C{l_2}}} = 1,6\,mol\\
{m_{Na}} = 1,6 \times 23 = 36,8g\\
b)\\
{n_{NaCl}} = 2{n_{C{l_2}}} = 1,6\,mol\\
{m_{NaCl}} = \dfrac{{1,6}}{{80\% }} \times 58,5 = 117g
\end{array}\)