Thí nghiệm 1 : $Al$ dư
$Ba + 2H_2O \to Ba(OH)_2 + H_2(1)$
$Ba(OH)_2 + 2Al + 2H_2O \to Ba(AlO_2)_2 + 3H_2(2)$
Thí nghiệm 2 :
$2Al + 2NaOH + 2H_2O \to 2NaAlO_2 + 3H_2$
Gọi $n_{Ba} = a(mol)$
Theo PTHH (1) : $n_{Ba(OH)_2} = n_{Ba} = a(mol)$
Theo PTHH (1)(2) : $n_{H_2} = n_{Ba} + 3n_{Ba(OH)_2} = 4a = 0,25$
$\Rightarrow a = 0,0625$
Theo PTHH (1)(2)(3) : $n_{H_2} = n_{Ba} + 3n_{Ba(OH)_2} + \dfrac{3}{2}n_{Al} = 0,4$
$\Rightarrow n_{Al} = 0,1$
Vậy : $n_{Al\ trong\ A} = 2n_{Ba(OH)_2} + n_{Al(3)} = 0,225(mol)$
$\Rightarrow m = 0,225.27 + 0,0625.137 = 14,6375(gam)$