\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Mg + 2HCl -> MgCl2 + H2 (1)
2Al + 6HCl -> 2AlCl3 + 3H2 (2)
Fe + 2HCl -> FeCl2 + H2 (3)
(1) ( 2)(3) => \(n_{HCl}=2n_{H_2}=0,6.2=1,2\left(mol\right)\)
\(m=25,12+1,2.36,5-0,6.2=67,72\left(g\right)\)
Đáp án A.