$n_K=\dfrac{97,5}{39}=2,5(mol)$
$n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2(mol)$
$2K+2HCl\to 2KCl+H_2$ $(1)$
$\dfrac{2,5}{1}<\dfrac{0,2}{1}\to K$ dư, $HCl$ hết
$\to K$ có phản ứng với nước.
$n_{K(1)}=n_{KCl}=n_{HCl}=0,2(mol)$
$\to n_{K(2)}=2,5-0,2=2,3(mol)$
$2K+2H_2O\to 2KOH+H_2$ $(2)$
$\to n_{KOH}=n_{K(2)}=2,3(mol)$
Ta có $n_{H_2}=n_{H_2(1)}+n_{H_2(2)}=\dfrac{1}{2}n_{K(1)}+\dfrac{1}{2}n_{K(2)}=\dfrac{1}{2}(n_{K(1)}+n_{K(2)})=\dfrac{n_K}{2}=1,25(mol)$
$\to V_{H_2}=1,25.22,4=28l$
$m_{dd\rm pứ}=m_K+100-m_{H_2}=97,5+100-1,25.2=195g$
$C\%_{KCl}=\dfrac{0,2.74,5.100}{195}=7,64\%$
$C\%_{KOH}=\dfrac{2,3.56.100}{195}=66,05\%$
