\(2Al+6HCl-->2AlCl4+3H2\)
x--------------------------------------1,5x(mol)
\(Fe+2HCl-->FeCl2+H2\)
y------------------------------------y(mol)
\(n_{H2}=\frac{10,08}{22,4}=0,45\left(mol\right)\)
Theo bài ra ta có hpt
\(\left\{{}\begin{matrix}27x+56y=13,8\\1,5x+y=0,45\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,15\end{matrix}\right.\)
\(\%m_{Al}=\frac{0,2.27}{13,8}.100\%=39,13\%\)
\(\%m_{Fe}=100-39,13=60,87\%\)
b) \(n_{HCl}=2n_{H2}=0,9\left(mol\right)\)
\(a=C_{M\left(HCl\right)}=\frac{0,9}{0,2}=0,45\left(M\right)\)
\(m_{muối}=m_{AlCl3}+m_{FeCl2}=0,2.133,5+0,15.127=45,75\left(g\right)\)
