a)
2KOH+ H2SO4---> K2SO4+2H20
mH2SO4=14,7. 0,2/100 =29,4 (g)
nH2SO4 =29,4/98= 0,3
mKOH=0,3.56=16,8
mdd KOH=16,8.100/5,6= 300
VKOH=300/10,45=28,7
b)
mK2SO4=0,3.174= 52,2g
mdd=28,7+200=228,7
C%= 52,2/228,7.100=22,82%
2KOH + H2SO4 → K2SO4 + 2H2O
\(m_{KOH}=200\times5,6\%=11,2\left(g\right)\)
\(\Rightarrow n_{KOH}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=0,2\times1,5=0,3\left(mol\right)\)
Theo PT: \(n_{KOH}=2n_{H_2SO_4}\)
Theo bài: \(n_{KOH}=\dfrac{2}{3}n_{H_2SO_4}\)
Vì \(\dfrac{2}{3}< 2\) ⇒ H2SO4 dư
Dung dịch sau phản ứng gồm: H2SO4 dư và K2SO4
Theo PT: \(n_{H_2SO_4}pư=\dfrac{1}{2}n_{KOH}=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}dư=0,3-0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}dư=0,2\times98=19,6\left(g\right)\)
Theo PT: \(n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
\(\Rightarrow m_{K_2SO_4}=0,1\times174=17,4\left(g\right)\)
\(m_{ddH_2SO_4}=200\times1,3=260\left(g\right)\)
\(m_{dd}saupư=m_{ddKOH}+m_{ddH_2SO_4}=200+260=460\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}dư=\dfrac{19,6}{460}\times100\%=4,26\%\)
\(C\%_{K_2SO_4}=\dfrac{17,4}{460}\times100\%=3,78\%\)
