a)
$2K + 2H_2O \to 2KOH + H_2$
$Ba + 2H_2O \to Ba(OH)_2 + H_2$
b)
n K = 7,8/39 = 0,2(mol)
n Ba = 41,1/137 = 0,3(mol)
Theo PTHH :
n H2 = 1/2 n K + n Ba = 0,4(mol)
=> V H2 = 0,4.22,4 = 8,96 lít
Theo gt ta có: $n_{K}=0,2(mol);n_{Ba}=0,3(mol)$
$2K+2H_2O\rightarrow 2KOH+H_2$
$Ba+2H_2O\rightarrow Ba(OH)_2+H_2$
Ta có: $n_{H_2}=0,4(mol)\Rightarrow V_{H_2}=8,96(l)$