a,
Đặt $2x$, $x$ là số mol $K$ và $Ca$
$n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)$
$2K+2H_2O\to 2KOH+H_2$
$Ca+2H_2O\to Ca(OH)_2+H_2$
Theo PTHH, $n_{H_2}=\dfrac{n_{Na}}{2}+n_{Ca}$
$\to 2x.\dfrac{1}{2}+x=0,4$
$\to x=0,2$
$\to n_K=0,4(mol); n_{Ca}=0,2(mol)$
$\to m=0,4.39+0,2.40=23,6g$
b,
Theo PTHH:
$n_{KOH}=2x=0,4(mol)$
$n_{Ca(OH)_2}=x=0,2(mol)$
$\to m_{\text{chất rắn}}=0,4.56+0,2.74=37,2g$