Đáp án:
\(\begin{array}{l}
\% {m_{Fe}} = 54,17\% \\
\% {m_{Mg}} = 26,79\% \\
\% {m_{Cu}} = 19,04\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
b)\\
hh:Fe(a\,mol),Mg(b\,mol),Cu(c\,mol)\\
{n_{{H_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35\,mol\\
\left\{ \begin{array}{l}
56a + 24b + 64c = 16,8\\
a + b = 0,35\\
64c = 0,32
\end{array} \right.\\
\Rightarrow a = 0,1625;b = 0,1875;c = 0,05\\
\% {m_{Fe}} = \dfrac{{0,1625 \times 56}}{{16,8}} \times 100\% = 54,17\% \\
\% {m_{Mg}} = \dfrac{{0,1875 \times 24}}{{16,8}} \times 100\% = 26,79\% \\
\% {m_{Cu}} = 100 - 54,17 - 26,79 = 19,04\%
\end{array}\)