Giải thích các bước giải:
Gọi a và b lần lượt là số mol của Fe và Cu
\(\begin{array}{l}
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 3{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
F{e_2}{(S{O_4})_3} + 6NaOH \to 2Fe{(OH)_3} + 3N{a_2}S{O_4}\\
CuS{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_2}S{O_4}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{n_{S{O_2}}} = 0,45mol\\
\left\{ \begin{array}{l}
\dfrac{3}{2}a + b = 0,45\\
160a + 80b = 28
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,2\\
b = 0,15
\end{array} \right.\\
\to {n_{Fe}} = 0,2mol \to {m_{Fe}} = 11,2g\\
\to {n_{Cu}} = 0,15mol \to {m_{Cu}} = 9,6g\\
a)\\
\% {m_{Fe}} = \dfrac{{11,2}}{{11,2 + 9,6}} \times 100\% = 53,85\% \\
\% {m_{Cu}} = 100\% - 53,85\% = 46,15\% \\
b)\\
{n_{{H_2}S{O_4}}} = 3{n_{Fe}} + 2{n_{Cu}} = 0,9mol \to {m_{{H_2}S{O_4}}} = 88,2g\\
\to C{\% _{{H_2}S{O_4}}} = \dfrac{{88,2}}{{99}} \times 100\% + 10\% = 89,2\% \\
c)\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Fe}} = 0,1mol \to {m_{F{e_2}{{(S{O_4})}_3}}} = 40g\\
{n_{CuS{O_4}}} = {n_{Cu}} = 0,15mol \to {m_{CuS{O_4}}} = 24g\\
{m_{{\rm{dd}}X}} = {m_{hh}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{S{O_2}}} = 91g\\
\to \% {C_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{40}}{{91}} \times 100\% = 44\% \\
\to \% {C_{CuS{O_4}}} = \dfrac{{24}}{{91}} \times 100\% = 26,4\%
\end{array}\)
