\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x_____________________3/2x
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
y ____________________y
Ta có :
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
Giải hệ phương trình :
\(\left\{{}\begin{matrix}27x+24y=6,3\\\frac{3}{2}x+y=0,3\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
\(m_{Al}=0,1.27=2,7\left(g\right)\)
\(m_{Mg}=0,15.24=3,6\left(g\right)\)
\(\%m_{Al}=\frac{2,7}{6,3}.100\%=42,86\left(g\right)\)
\(\%m_{Mg}=100\%-42,86\%=57,14\%\)
\(\rightarrow V_{dd_{HCl}}=\frac{0,6}{0,4}=1,5\left(l\right)\)
