$PTHH:2K+2H_2O→2KOH+H_2$ $(1)$
$K_2O+H_2O→2KOH$ $(2)$
$a.\ n_{H_2}=2,24:22,4=0,1mol$
Theo $PTHH\ (1):n_{K}=n_{KOH}=2n_{H_2}=0,2mol$
$→m_{K}=0,2.39=7,8g$
$\%K=\frac{7,8}{17,2}.100\%=45,35\%$
$\%K_2O=100\%-45,35\%=54,65\%$
$b.\ n_{K_2O}=\frac{17,2-7,8}{94}=0,1mol$
Theo $PTHH\ (2): n_{KOH}=2n_{K_2O}=0,2mol$
Tổng $n_{KOH}=0,2+0,2=0,4mol$
$→m_{KOH}=0,4.56=22,4g$
$m_{dd\ sp ư}=480g$
$→C\%_{KOH}=\frac{22,4}{480}.100\%=4,66\%$
$C_M=\frac{0,4}{0,4}=1M$