Giải thích các bước giải:
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{H_2}S{O_4} + Ba{(OH)_2} \to BaS{O_4} + 2{H_2}O(1)\\
A{l_2}{(S{O_4})_3} + 3Ba{(OH)_2} \to 3BaS{O_4} + 2Al{(OH)_3}(2)\\
FeS{O_4} + Ba{(OH)_2} \to BaS{O_4} + Fe{(OH)_2}(3)\\
2Al{(OH)_3} \to A{l_2}{O_3} + 3{H_2}O\\
Fe{(OH)_2} \to FeO + {H_2}O\\
{n_{{H_2}}} = 0,6mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{400 \times 19,6\% }}{{100\% }} = 78,4g\\
\to {n_{{H_2}S{O_4}}} = 0,8mol
\end{array}\)
Gọi a và b là số mol của Al và Fe
\(\begin{array}{l}
\left\{ \begin{array}{l}
27a + 56b = 22,2\\
\dfrac{3}{2}a + b = 0,6
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,2\\
b = 0,3
\end{array} \right.\\
\to {n_{Al}} = 0,2mol \to {m_{Al}} = 5,4g\\
\to {n_{Fe}} = 0,3mol \to {m_{Fe}} = 16,8g\\
\to \% {m_{Al}} = \dfrac{{5,4}}{{22,2}} \times 100\% = 24,32\% \\
\to \% {m_{Fe}} = \dfrac{{16,8}}{{22,2}} \times 100\% = 75,68\% \\
\to {n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} + {n_{Fe}} = 0,6mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,2mol\\
{n_{Ba{{(OH)}_2}}} = 0,84mol\\
\to {n_{Ba{{(OH)}_2}}}(1) = {n_{{H_2}S{O_4}(du)}} = 0,2mol\\
\to {n_{Ba{{(OH)}_2}}}(2 + 3) = 0,84 - 0,2 = 0,64mol\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al}} = 0,1mol\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,3mol\\
\to {n_{Ba{{(OH)}_2}}} = 3{n_{A{l_2}{{(S{O_4})}_3}}} + {n_{FeS{O_4}}} = 0,6mol\\
\to {n_{Ba{{(OH)}_2}}}dư = 0,04mol\\
\to {n_{Al{{(OH)}_3}}} = 2{n_{A{l_2}{{(S{O_4})}_3}}} = 0,2mol\\
\to {n_{A{l_2}{O_3}}} = \dfrac{1}{2}{n_{Al{{(OH)}_3}}} = 0,1mol\\
\to {n_{Fe{{(OH)}_2}}} = {n_{FeS{O_4}}} = 0,3mol\\
\to {n_{FeO}} = {n_{Fe{{(OH)}_2}}} = 0,3mol\\
\to {m_{chấtrắn}} = {m_{A{l_2}{O_3}}} + {m_{FeO}} = 0,1 \times 102 + 0,3 \times 72 = 31,8g
\end{array}\)
