$n_K=3,9/39=0,1mol$
$a.2K+2H_2O\to 2KOH+H_2↑$
b.Theo pt :
$n_{H_2}=1/2.n_K=1/2.0,1=0,05mol$
$⇒V_{H_2}=0,05.22,4=1,12l$
c.Theo pt :
$n_{KOH}=n_K=0,1mol$
$⇒C_{M_{KOH}}=\dfrac{0,1}{0,1}=0,5M$
$d.m_{KOH}=0,1.56=5,6g$
$m_{ddspu}=3,9+200-0,05.2=203,8g$
$⇒C\%_{KOH}=\dfrac{5,6}{203,8}=2,75\%$