Chất rắn không tan là Cu không tham gia phản ứng => mCu = 2,5 (g)
=> mMg + mAl = 12,7 - 2,5 = 10,2 (g)
Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\left(ĐK:x,y>0\right)\)
=> 24x + 27y = 10,2 (1)
PTHH:
Mg + 2HCl -> MgCl2 + H2
x--------------------------->x
2Al + 6HCl -> 2AlCl3 + 3H2
y--------------------------->1,5y
Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) => x + 1,5y = 0,5 (2)
Từ (1), (2) => x = y = 0,2 (mol) (t/m)
=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{2,5}{12,7}.100\%=19,69\%\\\%m_{Al}=\dfrac{0,2.27}{12,7}.100\%=42,52\%\\\%m_{Mg}=100\%-19,69\%-42,52\%=37,79\%\end{matrix}\right.\)
