Đáp án:
Bạn tham khảo lời giải ở dưới nhé !
Giải thích các bước giải:
\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 9,8\% }}{{100\% }} = 19,6g\\
\to {n_{{H_2}S{O_4}}} = 0,2mol
\end{array}\)
Gọi a và b lần lượt là số mol của Fe và Al
\(\begin{array}{l}
\left\{ \begin{array}{l}
56a + 27b = 5,55\\
a + \dfrac{3}{2}b = 0,2
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,075\\
b = 0,05
\end{array} \right.\\
\to {n_{Fe}} = 0,075mol\\
\to {n_{Al}} = 0,05mol\\
a)\\
\% {m_{Fe}} = \dfrac{{0,075 \times 56}}{{5,55}} \times 100\% = 75,68\% \\
\% {m_{Al}} = 100\% - 75,68\% = 24,32\%
\end{array}\)
\(\begin{array}{l}
b)\\
{H_2}S{O_4} + Ba{(OH)_2} \to B{\rm{aS}}{O_4} + 2{H_2}O\\
FeS{O_4} + Ba{(OH)_2} \to Fe{(OH)_2} + B{\rm{aS}}{O_4}\\
A{l_2}{(S{O_4})_3} + 3Ba{(OH)_2} \to 2Al{(OH)_3} + 3B{\rm{aS}}{O_4}\\
{n_{{H_2}S{O_4}(a)}} = {n_{Fe}} + \dfrac{3}{2}{n_{Al}} = 0,15mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,05mol\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,075mol\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al}} = 0,025mol\\
{n_{Ba{{(OH)}_2}}} = 0,21mol\\
\to {n_{Ba{{(OH)}_2}(pt)}} = {n_{{H_2}S{O_4}(du)}} + {n_{FeS{O_4}}} + 3{n_{A{l_2}{{(S{O_4})}_3}}} = 0,2mol\\
\to {n_{Ba{{(OH)}_2}(dư)}} = 0,01mol\\
Ba{(OH)_2} + 2Al{(OH)_3} \to Ba{(Al{O_2})_2} + 4{H_2}O\\
{n_{Al{{(OH)}_3}}} = 2{n_{A{l_2}{{(S{O_4})}_3}}} = 0,05mol\\
\to \dfrac{{{n_{Ba{{(OH)}_2}(dư)}}}}{1} < \dfrac{{{n_{Al{{(OH)}_3}}}}}{2} \to {n_{Al{{(OH)}_3}}}dư\\
\to {n_{Al{{(OH)}_3}(pt)}} = 2{n_{Ba{{(OH)}_2}(dư)}} = 0,02mol\\
\to {n_{Al{{(OH)}_3}(dư)}} = 0,03mol\\
4Fe{(OH)_2} + {O_2} \to 2F{e_2}{O_3} + 4{H_2}O\\
2Al{(OH)_3} \to A{l_2}{O_3} + 3{H_2}O\\
{n_{Fe{{(OH)}_2}}} = {n_{FeS{O_4}}} = 0,075mol\\
\to {n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe{{(OH)}_2}}} = 0,0375mol\\
\to {m_{F{e_2}{O_3}}} = 6g\\
{n_{A{l_2}{O_3}}} = \dfrac{1}{2}{n_{Al{{(OH)}_3}(dư)}} = 0,015mol\\
\to {m_{A{l_2}{O_3}}} = 1,53g\\
{n_{{\rm{BaS}}{O_4}}} = {n_{{H_2}S{O_4}(dư)}} + {n_{FeS{O_4}}} + 3{n_{A{l_2}{{(S{O_4})}_3}}} = 0,2mol\\
\to {m_{{\rm{BaS}}{O_4}}} = 46,6g
\end{array}\)
