(Sửa Ca thành Cu)
Gọi a, b, c là mol Al, Cu, Mg.
$\Rightarrow 27a+64b+24c=10$ (1)
$n_{H_2}=0,534 mol$
$2Al+6HCl\to 2AlCl_3+3H_2$
$Mg+2HCl\to MgCl_2+H_2$
$\Rightarrow 1,5a+c=0,534$ (2)
Chất không tan là Cu.
$\Rightarrow 64b=0,31$ (3)
(1)(2)(3) $\Rightarrow a=0,347; b=0,0048; c=0,013$
$\%m_{Al}=\frac{0,347.27.100}{10}=93,69\%$
$\%m_{Mg}=\frac{0,31.100}{10}=3,1\%$
$\%m_{Cu}=3,21\%$