nH2= \(\frac{1,456}{22,4}\)= 0,065 mol
Gọi nAl = a (mol); nFe = b (mol)
=> 27a + 56b = 1,93 (I)
PTHH: 2Al + 6HCl→ 2AlCl3 + 3H2 (1)
_______x----------------->x----->1,5x______(mol)
Fe + 2HCl→ FeCl2 + H2 (2)
y---------------->y----->y_________(mol)
=> 1,5x + y = 0,065 (mol) (II)
(I)(II) => \(\left\{{}\begin{matrix}x=0,03\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)
=> \(m=0,03.133,5+0,02.127=6,545\left(g\right)\)