4g chất rắn không tan là Cu
\(\Rightarrow m_{Mg}+Al=13-4=9\left(g\right)\)
\(n_{H2}=\frac{10,08}{22,4}=0,45\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x____________________1,5x
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
y_____________________y
Gọi x, y lần lượt là nAl và nMg, ta có:
Giải hệ PT:
\(\left\{{}\begin{matrix}27x+24y=9\\1,5x+y=0,45\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,15\end{matrix}\right.\)
\(\Rightarrow m_{A;}=0,2.27=5,4\left(g\right)\)
\(\Rightarrow\%m_{Al}=\frac{5,4}{13}.100\%=41,54\%\)
\(m_{Mg}=0,15.24=3,6\left(g\right)\)
\(\Rightarrow\%m_{Mg}=\frac{3,6}{13}.100\%=27,7\%\)
\(\Rightarrow\%m_{Cu}=100\%-\left(41,54+27,7\right)=30,76\%\)
