$PTHH : \\Mg+2HCl\to MgCl_2+H_2↑ \\2Al+6HCl\to 2AlCl_3+3H_2↑$

Chất rắn không tan sau phản ứng là Cu

$⇒m_{Cu}=6,4g \\n_{H_2}=\dfrac{13,44}{22,4}=0,6mol \\Gọi\ n_{Mg}=a;n_{Al}=b \\Ta\ có :$ $m_{hh}=m_{Mg}+m_{Al}+6,4=24a+27b+6,4=19(g) \\n_{H_2}=n_{Mg}+\dfrac{3}{2}.n_{Al}=a+1,5b=0,6mol \\Ta\ có\ hpt :$ $\left\{\begin{matrix} 24a+27b+6,4=19 & \\  a+1,5b=0,6 &  \end{matrix}\right.⇔\left\{\begin{matrix} a=0,3 & \\  b=0,2 &  \end{matrix}\right.$ $⇒\%m_{Mg}=\dfrac{0,3.24}{19}.100\%=37,89\% \\\%m_{Al}=\dfrac{0,2.27}{19}.100\%=28,42\% \\\%m_{Cu}=100\%-37,89\%-28,42\%=33,69\%$  

$Mg + 2HCl ---> MgCl_2 + H_2$

$2Al + 6HCl ---> 2AlCl_3 + 3H_2$

$n_{H_2}= \dfrac{13,44}{22,4}= 0,6mol$

Gọi x là mol Mg; y là mol Al

Ta có hpt:

$\left \{ {{x+3/2y=0,6} \atop {24x+ 27y=19-6,4}} \right.$

$\left \{ {{x=0,3} \atop {y=0,2}} \right.$

=> $m_{Mg}= 0,3.24= 7,2g; m_{Al}= 0,2.27= 5,4g$

Chất rắn không tan là Cu

%$m_{Mg/X}= \dfrac{7,2}{19}= 37,8$%

%$m_{Al/X}= \dfrac{5,4}{19}= 28,4$%

%$m_{Cu/X}= \dfrac{6,4}{19}= 33,8$%