P1:
4Al + 3O2 -> 2Al2O3
2a a
2Ba + O2 -> 2BaO
2b 2b
P2:
2Al + 6HCl -> 2AlCl3 + 3H2
2a 3a
Ba + 2HCl -> BaCl2 + H2
2b 2b
$n_{H2}$=$\frac{8,96}{22,4}$=0,4 (mol)
PtP1: 102a+306b=25,5
PtP2: 3a+2b=0,4
⇒$\left \{ {{x=0,1} \atop {y=0,05}} \right.$
⇒$m_{Al}$=27.0,1.2.2=10,8 (g)
$m_{Ba}$=137.0,05.2.2=27,4 (g)