Em tham khảo nha:

\(\begin{array}{l}
3)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{3,42}}{{342}} = 0,01\,mol\\
{n_{Al{{(OH)}_3}}} = \dfrac{{1,56}}{{78}} = 0,02\,mol\\
2{n_{A{l_2}{{(S{O_4})}_3}}} = {n_{Al{{(OH)}_3}}} \Rightarrow\text{ Phản ứng vừa đủ, kết tủa không tan}\\
A{l_2}{(S{O_4})_3} + 6NaOH \to 2Al{(OH)_3} + 3N{a_2}S{O_4}\\
{n_{NaOH}} = 6{n_{A{l_2}{{(S{O_4})}_3}}} = 0,01 \times 6 = 0,06\,mol\\
{C_M}NaOH = \dfrac{{0,06}}{{0,05}} = 1,2M\\
4)\\
{n_{AlC{l_3}}} = 0,2 \times 1 = 0,2\,mol\\
{n_{Al{{(OH)}_3}}} = \dfrac{{7,8}}{{78}} = 0,1\,mol\\
{n_{AlC{l_3}}} > {n_{Al{{(OH)}_3}}} \Rightarrow\text{ $AlCl_3$ dư hoặc kết tủa tan 1 phần} \\
T{H_1}:\text{$AlCl_3$ dư}\\
AlC{l_3} + 3KOH \to Al{(OH)_3} + 3KCl\\
{n_{NaOH}} = 3{n_{Al{{(OH)}_3}}} = 0,1 \times 3 = 0,3\,mol\\
{C_M}KOH = \dfrac{{0,3}}{{0,2}} = 1,5M\\
T{H_2}:\text{ Kết tủa tan 1 phần}\\
AlC{l_3} + 3KOH \to Al{(OH)_3} + 3KCl(1)\\
Al{(OH)_3} + KOH \to KAl{O_2} + 2{H_2}O(2)\\
{n_{Al{{(OH)}_3}(1)}} = {n_{AlC{l_3}}} = 0,2\,mol\\
{n_{Al{{(OH)}_3}(2)}} = 0,2 - 0,1 = 0,1\,mol\\
{n_{KOH}} = 0,2 \times 3 + 0,1 = 0,7\,mol\\
{C_M}KOH = \dfrac{{0,7}}{{0,2}} = 3,5M
\end{array}\)

 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`