a,
Gọi a, b là số mol K, Ba.
$\Rightarrow 39x+137y=4,3$ (1)
$n_{H_2}=\dfrac{0,896}{22,4}=0,04(mol)$
$2K+2H_2O\to 2KOH+H_2$
$Ba+2H_2O\to Ba(OH)_2+H_2$
$\Rightarrow 0,5a+b=0,04$ (2)
$(1)(2)\Rightarrow a=0,04; b=0,02$
$\%n_K=\dfrac{a.100}{a+b}=66,67\%$
$\%n_{Ba}=33,33\%$