Đáp án:
\(\begin{array}{l}
\% {m_{Fe}} = 2,11\% \\
\% {m_{Zn}} = 97,89\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Zn + 2{H_2}S{O_4} \to ZnS{O_4} + S{O_2} + 2{H_2}O\\
b)\\
hh:Fe(a\,mol),Zn(b\,mol)\\
{n_{S{O_2}}} = \dfrac{3}{2} \times {n_{Fe}} + {n_{Zn}} = 1,5a + b\,mol\\
{m_{{\rm{dd}}spu}} = {m_{{\rm{dd}}{H_2}S{O_4}}} \Rightarrow {m_{hh}} - {m_{S{O_2}}} = 0\\
\Leftrightarrow 56a + 65b = 96a + 64b \Leftrightarrow 40a = b\\
\% {m_{Fe}} = \dfrac{{a \times 56}}{{a \times 56 + b \times 65}} \times 100\% = \dfrac{{56a}}{{56a + 40a \times 65}} \times 100\% = 2,11\% \\
\% {m_{Zn}} = 100 - 2,11 = 97,89\%
\end{array}\)