Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
4Fe{(OH)_2} + {O_2} + 2{H_2}O \to 4Fe{(OH)_3}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{{16}}{{160}} = 0,1mol\\
{n_{Fe{{(OH)}_3}}} = 2{n_{F{e_2}{O_3}}} = 0,2mol\\
{n_{Fe{{(OH)}_2}}} = {n_{Fe{{(OH)}_3}}} = 0,2mol\\
{n_{NaOH}} = 2{n_{Fe{{(OH)}_2}}} = 0,4mol\\
{n_{Na}} = {n_{NaOH}} = 0,4mol\\
{m_{Na}} = 0,4 \times 23 = 9,2g\\
{n_{{H_2}}} = \dfrac{{{n_{Na}}}}{2} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
{m_{FeC{l_2}}} = \dfrac{{63,5 \times 100}}{{100}} = 63,5g\\
{n_{FeC{l_2}}} = \dfrac{{63,5}}{{127}} = 0,5mol\\
{n_{FeC{l_2}d}} = 0,5 - 0,2 = 0,3mol\\
{m_{FeC{l_2}d}} = 0,3 \times 127 = 38,1g\\
{m_{Fe{{(OH)}_2}}} = 0,2 \times 90 = 18g\\
{m_{ddspu}} = 9,2 + 100 - 18 - 0,2 \times 2 = 90,8g\\
{n_{NaCl}} = 2{n_{Fe{{(OH)}_2}}} = 0,4mol\\
{m_{NaCl}} = 0,4 \times 58,5 = 23,4g\\
C{\% _{FeC{l_2}d}} = \dfrac{{38,1}}{{90,8}} \times 100\% = 42\% \\
C{\% _{NaCl}} = \dfrac{{23,4}}{{90,8}} \times 100\% = 25,77\%
\end{array}\)
