Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 10,24\% \\
\% {m_{Cu}} = 89,76\% \\
b)\\
m = 85,6g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{{12,32}}{{22,4}} = 0,55\,mol\\
hh:Fe(a\,mol),Cu(b\,mol)\\
\left\{ \begin{array}{l}
56a + 64b = 32,8\\
1,5a + b = 0,55
\end{array} \right.\\
 \Rightarrow a = 0,06;b = 0,46\,mol\\
\% {m_{Fe}} = \dfrac{{0,06 \times 56}}{{32,8}} \times 100\%  = 10,24\% \\
\% {m_{Cu}} = 100 - 10,24 = 89,76\% \\
b)\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \frac{{0,06}}{2} = 0,03\,mol\\
{n_{CuS{O_4}}} = {n_{Cu}} = 0,46\,mol\\
m = 0,03 \times 400 + 0,46 \times 160 = 85,6g
\end{array}\)