\(\begin{array}{l}
5)\\
(I):\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
(II):\\
Mg + C{l_2} \to MgC{l_2}\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
Cu + C{l_2} \to CuC{l_2}\\
hh:Mg(a\,mol),Fe(b\,mol),Cu(c\,mol)\\
n{H_2} = \dfrac{{3,808}}{{22,4}} = 0,17\,mol\\
nC{l_2} = \dfrac{{5,712}}{{22,4}} = 0,255\,mol\\
24a + 56b + 64c = 9,52\\
0,5a + 0,5b = 0,17\\
0,5a + 0,75b + 0,5c = 0,255\\
\Rightarrow a = b = c =
\end{array}\)
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