$PTHH:2K+2H_2O→2KOH+H_2$ $(1)$

$K_2O+H_2O→2KOH$ $(2)$

$a.\ n_{H_2}=2,24:22,4=0,1mol$

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 $n_{KOH}=0,2+0,2=0,4mol$

$→m_{KOH}=0,4.56=22,4g$

$m_{dd\ sp ư}=480g$

$→C\%_{KOH}=\frac{22,4}{480}.100\%=4,66\%$

$C_M=\frac{0,4}{0,4}=1M$

ta có phương trình:

2K+2H2O+>2KOH+H2↑

 K2O+H2O=>2KOH

a,ta có vH2=2.24(lít)=>nH2=$\frac{2.24}{22.4}$=0.1(mol) 

=>nK=0.1*2=0.2(mol)

=>mK=0.2*39=7.8(g)

=>%K=$\frac{7.8}{17.2}$*100= 45.35%

=>%K2O=$\frac{17.2-7.8}{17.2}$*100=54.65% 

b,ta có nK2O=$\frac{17.2-7.8}{94}$=0.1(mol) =>nKOH=0.1*2=0.2(mol)=>mKOH=0.2*56=11.2(g)

mKOH=(0.1*2)*56=11.2(g)

=>mA=11.2+11.2=22.4(g);nA=$\frac{22.4}{56}$=0.4(mol) 

ta có m=D.v=1.2*400=480(g)

=>%A=$\frac{22.4}{480}$*100=4.67% 

=>CmA=$\frac{0.4}{0.4}$=1M