a,
$n_K=\dfrac{3,9}{39}=0,1(mol)$
$2K+2H_2O\to 2KOH+H_2$
Theo PTHH:
$n_{H_2}=\dfrac{n_K}{2}=\dfrac{0,1}{2}=0,05(mol)$
$\to V_{H_2}=0,05.22,4=1,12l$
b,
$m_{dd\text{spứ}}=m_K+m_{H_2O}-m_{H_2}= 3,9+36,2-0,05.2=40g$
Theo PTHH:
$n_{KOH}=n_K=0,1(mol)$
$\to m_{KOH}=0,1.56=5,6g$
$\to C\%_{KOH}=\dfrac{5,6.100}{40}=14\%$