Bài 3
`n_(Al_2(SO_4)_3)=\frac{3,42}{342}=0,01(mol)`
`n_(Al(OH)_3)=\frac{1,56}{78}=0,02(mol)`
Ta có
`n_(Al (Al_2(SO_4)_3))=n_(Al (Al(OH)_3)=0,02(mol)`
`Al_2(SO_4)_3+6NaOH->3Na_2SO_4+2Al(OH)_3`
Theo PT
`n_(NaOH)=6n_(Al_2(SO_4)_3)=0,06(mol)`
`=>C_(MNaOH)=\frac{0,06}{0,05}=1,2M`
Bài 4
`n_(AlCl_3)=0,2.1=0,2(mol)`
`n_(Al(OH)_3)=\frac{7,8}{78}=0,1(mol)`
Ta có
`n_(Al (AlCl_3))>n_(Al (Al(OH)_3)`
`=>AlCl_3` dư hoặc `Al(OH)_3` tan 1 phần
`TH1:AlCl_3`
`AlCl_3+3KOH->Al(OH)_3+3KCl`
Theo PT
`n_(KOH)=3n_(Al(OH)_3)=0,3(mol)`
`=>C_(MKOH)=\frac{0,3}{0,2}=1,5M`
`TH2:Al(OH)_3` tan 1 phần
`AlCl_3+3KOH->Al(OH)_3+3KCl (1)`
`Al(OH)_3+KOH->KAlO_2+2H_2O (2)`
`n_(Al(OH)_3 (1))=n_(AlCl_3)=0,2(mol)`
`n_(Al(OH)_3 (2))=n_(Al(OH)_3 (1))-n_(Al(OH)_3)=0,2-0,1=0,1(mol)`
Theo PT
`n_(KOH)=3n_(AlCl_3)+n_(Al(OH)_3 (2))=3.0,2+0,1=0,7(mol)`
`=>C_(MKOH)=\frac{0,7}{0,2}=3,5M`
