`1)`

Dùng nước:

+ Tan, sủi bọt khí: K

+ Không tan: Al,Mg (1)

Cho dd `KOH` vào (1):

+ Tan, sủi bọt khí: Al

+ Không tan: Mg

`2K+2H_2O->2KOH+H_2`

`2Al+2KOH+2H_2O->2KAlO_2+3H_2`

`2)`

`2Fe+3Cl_2` $\xrightarrow{t^o}$ `2FeCl_3`

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

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

`Fe_2O_3+3H_2SO_4->Fe_2(SO_4)_3+3H_2O`

`3)`

`a)`

`Fe+2HCl->FeCl_2+H_2`

`Fe_2O_3+6HCl->2FeCl_3+3H_2O`

`b)`

Theo pt:

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

`\%m_{Fe}={0,3.56}/{20,1}.100\%=83,58\%`

`\%m_{Fe_2O_3}=100-83,58=16,42\%`

`c)`

`n_{Fe_2O_3}={20,1-0,3.56}/{160}=0,020625(mol)`

Theo pt:

`n_{HCl}=2n_{Fe}+6n_{Fe_2O_3}=0,72325(mol)`

`V_{dd\ HCl}={0,72325}/2=0,361875(L)`

`d)`

`n_{FeCl_2}=n_{Fe}=0,3(mol);n_{FeCl_3}=2n_{Fe_2O_3}=0,04125(mol)`

`FeCl_2+2KOH->Fe(OH)_2+2KCl`

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

`->n_{KCl}=2n_{FeCl_2}+3n_{FeCl_3}=0,72325(mol)`

`m_{KCl}=0,72325.58,5=42,310125(g)`