`n_{K_2O}=(14,1)/94=0,15(mol)`
`K_2O+H_2O->2KOH`
`n_{KOH}=2n_{K_2O}=0,3(mol)`
`C%_{ddKOH}=(0,3.56)/100 .100=16,8%`
`KOH+HCl->KCl+H_2O`
`n_{HCl}=0,2.2=0,4(mol)`
`(0,3)/1<(0,4)/1=>HCl` dư
`V_{ddspư}=V_{ddHCl}=0,2(l)`
`n_{HClpư}=n_{KOH}=0,3(mol)`
`n_{HCldư}=0,4-0,3=0,1(mol)`
`n_{KCl}=n_{KOH}=0,3(mol)`
`C_{M_{HCl}}=(0,1)/(0,2)=0,5(M)`
`C_{M_{KCl}}=(0,3)/(0,2)=1,5(M)`
`\text{___________________________________}`
Đáp án:
`C%_{ddKOH}=16,8%`
`C_{M_{HCl}}=0,5(M)`
`C_{M_{KCl}}=1,5(M)`
