`D1:`

`1)`

Dùng dd NaOH:

+ Tan, sủi khí: Al

+ Không tan: Cu,Fe(1)

Cho dd HCl vào (1):

+ Tan: Fe

+ Không tan: Cu

`2Al+2NaOH+2H_2O->2NaAlO_2+3H_2`

`Fe+2HCl->FeCl_2+H_2`

`2)`

Dùng dd NaOH:

+ Tan, sủi khí: Al

+ Không tan: Ag,Fe(1)

Cho dd HCl vào (1):

+ Tan: Fe

+ Không tan: Ag

`2Al+2NaOH+2H_2O->2NaAlO_2+3H_2`

`Fe+2HCl->FeCl_2+H_2`

`D2:`

`1)`

`a)`

`Fe+2HCl->FeCl_2+H_2`

`b)`

`A:FeCl_2,HCl` dư

`B:Cu`

`c)`

`n_{Fe}=n_{H_2}={6,72}/{22,4}=0,3(mol)`

`\%m_{Fe}={0,3.56}/{30}.100\%=56\%`

`\%m_{Cu}=100-56=44\%`

`D3:`

`1)`

`FeCl_3+3KOH->Fe(OH)_3+3KCl`

`2Fe(OH)_3` $\xrightarrow{t^o}$ `Fe_2O_3+3H_2O`

`3Fe_2O_3+CO` $\xrightarrow{t^o}$ `2Fe_3O_4+CO_2`

`2)`

`4Al+3O_2` $\xrightarrow{t^o}$ `2Al_2O_3`

`Al_2O_3+6HCl->2AlCl_3+3H_2O`

`AlCl_3+3KOH->Al(OH)_3+3KCl`

`2Al(OH)_3` $\xrightarrow{t^o}$ `Al_2O_3+3H_2O`

`2Al_2O_3` $\xrightarrow{dpnc,criolit}$ `4Al+3O_2`

`3)`

`Fe+H_2SO_4->FeSO_4+H_2`

`FeSO_4+2KOH->Fe(OH)_2+K_2SO_4`

`Fe(OH)_2+2HCl->FeCl_2+2H_2O`