$PTHH :$
$2K+2H_2O\to 2KOH+H_2↑(1)$
$Ba+2H2O\to Ba(OH)_2+H_2↑(2)$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2mol$
a.Gọi $n_{K}=a(mol);n_{Ba}=b(mol) (a,b>0)$
Ta có :
$m_{hh}=39a+137b=21,5g$
$n_{H_2}=0,5a+b=0,2mol$
Ta có hpt :
$\left\{\begin{matrix}
39a+137b=21,5 & \\
0,5a+b=0,2 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,2 & \\
b=0,1 &
\end{matrix}\right.$
$⇒\%m_K=\dfrac{0,2.39}{21,5}.100\%=36,28\%$
$\%m_{Ba}=100\%-36,28\%=63,72\%$
b.Theo pt (1) và (2) :
$n_{KOH}=n_K=0,2mol$
$n_{Ba(OH)_2}=n_{Ba}=0,1mol$
$⇒C_{M_{KOH}}=\dfrac{0,2}{0,2}=1M$
$C_{M_{Ba(OH)_2}}=\dfrac{0,1}{0,2}=0,5M$