2Al + 6HCl ---> 2AlCl3 + 3H2
x -------------> 3/2 x (mol)
Fe + 2HCl ----> FeCl2 + H2
y --------------> y (mol)
=> \(\left\{{}\begin{matrix}27x+56y=5,5\\3x+2y=0,4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
=> %mAl = 0,1.27.100%/5,5= 49,09%
%mFe = 100% - 49,09% = 50,91%