TN1 lượng hidro thu được ít hơn TN2 nên TN1, dư Al.
Gọi x, y là số mol Ba, Al.
- TN1:
$n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$Ba+2H_2O\to Ba(OH)_2+H_2$
$2Al+Ba(OH)_2+2H_2O\to Ba(AlO_2)_2+3H_2$
$\Rightarrow n_{H_2}=n_{Ba}+1,5n_{Ba(OH)_2}=2,5n_{Ba}$
$\Rightarrow 2,5x=0,2$ (1)
- TN2:
$n_{H_2}=\dfrac{13,44}{22,4}=0,6(mol)$
$Ba+2H_2O\to Ba(OH)_2+H_2$
$2Al+Ba(OH)_2+2H_2O\to Ba(AlO_2)_2+3H_2$
$\Rightarrow x+1,5y=0,6$ (2)
(1)(2)$\Rightarrow x=0,08; y=\dfrac{26}{75}$
$m_{Al}=27x=2,16g$
$m_{Ba}=137y=47,49g$