Giải thích các bước giải:
\(\begin{array}{l}
2.\\
6Na + 6{H_2}O + F{e_2}{(S{O_4})_3} \to 2Fe{(OH)_3} + 3N{a_2}S{O_4} + 3{H_2}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
{n_{Na}} = 0,2mol\\
{m_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{57 \times 40\% }}{{100\% }} = 22,8g\\
\to {n_{F{e_2}{{(S{O_4})}_3}}} = 0,057mol\\
\to {n_{Na}} > {n_{F{e_2}{{(S{O_4})}_3}}} \to {n_{Na}}dư\\
\to {n_{Fe{{(OH)}_3}}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = 0,114mol\\
\to {n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe{{(OH)}_3}}} = 0,057mol\\
\to {m_{F{e_2}{O_3}}} = 9,12g
\end{array}\)
\(\begin{array}{l}
3\\
2Al{(OH)_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 6{H_2}O\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
{n_{{H_2}S{O_4}}} = 0,25
\end{array}\)
Gọi a và b là số mol của Al(OH)3 và MgO
\(\begin{array}{l}
\left\{ \begin{array}{l}
78a + 40b = 11,8\\
\dfrac{3}{2}a + b = 0,25
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,1\\
b = 0,1
\end{array} \right.\\
\to {n_{Al{{(OH)}_3}}} = 0,1mol \to {m_{Al{{(OH)}_3}}} = 7,8g\\
\to {n_{MgO}} = 0,1mol \to {m_{MgO}} = 4g\\
a)\\
\% {m_{Al{{(OH)}_3}}} = \dfrac{{7,8}}{{11,8}} \times 100\% = 66,1\% \\
\% {m_{MgO}} = \dfrac{4}{{11,8}} \times 100\% = 33,9\% \\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al{{(OH)}_3}}} = 0,05mol\\
{n_{MgS{O_4}}} = {n_{MgO}} = 0,1mol\\
\to {m_{muối}} = {m_{A{l_2}{{(S{O_4})}_3}}} + {m_{MgS{O_4}}} = 46,5g
\end{array}\)
