Đáp án:
49% ; 28,89% và 22,11%
Giải thích các bước giải:
\(\begin{array}{l}
{m_{Cu}} = m = 9,8g\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 10,2\\
1,5a + b = 0,4
\end{array} \right.\\
\Rightarrow a = \dfrac{{61}}{{285}};b = \dfrac{3}{{38}}\\
\% {m_{Cu}} = \dfrac{{9,8}}{{20}} \times 100\% = 49\% \\
\% {m_{Al}} = \dfrac{{\frac{{61}}{{285}} \times 27}}{{20}} \times 100\% = 28,89\% \\
\% {m_{Fe}} = 100 - 49 - 28,89 = 22,11\%
\end{array}\)