2Al+6HCl->2AlCl3+3H2
x----------------------------3\2x
Fe+2HCl->FeCl2+H2
y-------------------------y
=>\(\left\{{}\begin{matrix}27x+56y=21,1\\3\backslash2x+y=\dfrac{14,56}{22,4}\end{matrix}\right.\)
=>x=0,268 mol
y=0,247 mol
=>%m Al=\(\dfrac{0,268.27}{21,1}\).100=34,2938%
=>%m Fe=100-34,2938=65,7062