Đáp án:
a) $KOH + HCl \to KCl + H_2O$
b)$C_{KOH}\%=12\%$
c) $m_{CaO}= 1,4g$
d)
$C_{K_2SO_4}\%=13,51\%$
$C_{CaSO_4}\%=17,6\%$
Giải thích các bước giải:
$n_{HCl}= 0,2.0,75=0,15mol$
Phương trình phản ứng:
$KOH + HCl \to KCl + H_2O$
$0,15$ ←$0,15$ (mol)
$m_{KOH}= 0,15.56=8,4g$
$C_{KOH}\%=\frac{8,4.100}{70}=12\%$
$n_{H_2SO_4}= 0,2.0,2=0,04mol$
$m_{KOH}=\frac{14.12}{100}=1,68g$
$n_{KOH}=\frac{1,68}{56}=0,03mol$
$CaO + H_2O \to Ca(OH)_2$
$2KOH + H_2SO_4 \to K_2SO_4 + 2H_2O$
$0,03$→ $0,015$ $0,015$ (mol)
$n_{H_2SO_4 pư với Ca(OH)_2}= 0,04- 0,015=0,025mol$
$Ca(OH)_2 + H_2SO_4 \to CaSO_4 + 2H_2O$
$0,025$ ←$0,025$ → $0,025$ (mol)
$n_{Ca(OH)_2} =n_{CaO}= 0,025mol$
$m_{CaO}= 0,025.56=1,4g$
$m_{dung dịch}= 1,4 + 14+ 0,04.98=19,32g$
$m_{K_2SO_4}=0,015.174=2,61g$
$m_{CaSO_4}= 0,025.136=3,4g$
$C_{K_2SO_4}\%=\frac{2,61.100}{19,32}=13,51\%$
$C_{CaSO_4}\%=\frac{3,4.100}{19,32}=17,6\%$