\(\begin{array}{l}
1)\\
mNaCl\,trong\,{\rm{dd}}\,20\%  = 150 \times 20\%  = 30g\\
m{H_2}O\,trong\,{\rm{dd}}\,20\%  = 150 - 30 = 120g\\
mNaCl = \dfrac{{120 \times 36}}{{100}} = 43,2g\\
mNaCl = 43,2 - 30 = 13,2g\\
2)\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
nCa = \dfrac{m}{M} = \dfrac{4}{{40}} = 0,1\,mol\\
n{H_2} = nCa = 0,1\,mol\\
m{\rm{dd}} = 4 + 50 - 0,1 \times 2 = 53,8g\\
3)\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
nK = \dfrac{m}{M} = \dfrac{{3,2}}{{39}} = 0,082\,mol\\
n{H_2} = \dfrac{{0,082}}{2} = 0,041\,mol\\
m{\rm{dd}}spu = 3,2 + 20 - 0,041 \times 2 = 23,118g\\
nKOH = nK = 0,082\,mol\\
C\% KOH = \dfrac{{0,082 \times 56}}{{23,118}} \times 100\%  = 19,86\% 
\end{array}\)

Đáp án:

Giải thích các bước giải:

Bài 1: mNaCltrongdd20%=150×20%=30gmH2Otrongdd20%=150−30=120gmNaCl=120×36100=43,2gmNaCl=43,2−30=13,2g

bài2

Ca+2H2O→Ca(OH)2+H2nCa=mM=440=0,1molnH2=nCa=0,1molmdd=4+50−0,1×2=53,8g

bài 3 2K+2H2O→2KOH+H2nK=mM=3,239=0,082molnH2=0,0822=0,041molmddspu=3,2+20−0,041×2=23,118gnKOH=nK=0,082molC%KOH=0,082×5623,118×100%=19,86%

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