`a)`

`FeO+ 2HCl→ FeCl_2+ H_2O` 

`Fe_2O_3 + 6HCl→ 2FeCl_3 + 3H_2O`

`Fe_3O_4 + 8HCl→ 2FeCl_3 + FeCl_2 + 4H_2O` 

Vì không rõ số mol `FeCl_3` nên chỉ dựa theo quy tắc `alpha` đồng thời `HCl` dư nên ta có: 

`Mg + 2FeCl_3 → MgCl_2 + 2FeCl_2`

`Mg + 2HCl→ MgCl_2+ H_2`

`Fe + 2FeCl_3 → 3FeCl_2`

`Fe + 2HCl→ FeCl_2 + H_2` 

`C: FeCl_2, MgCl_2, HCl`

`NaOH+ HCl→ NaCl+ H_2O`

`FeCl_2 + 2NaOH→ Fe(OH)_2+ 2NaCl`

`MgCl_2+ 2NaOH→ Mg(OH)_2+ 2NaCl`

`Mg(OH)_2`$\xrightarrow{t°}$`MgO + H_2O`

`4Fe(OH)_2 + O_2`$\xrightarrow{t°}$`2Fe_2O_3 + 4H_2O`

`Fe_2O_3+ 3H_2`$\xrightarrow{t°}$`2Fe +3H_2O`

`b)` 

`m_{Fe} : m_{Mg}= 5:3`

`<=> 56n_{Fe} : 24n_{Mg} = 5:3`

`<=> 168n_{Fe} = 120n_{Mg}`

`<=> {n_{Mg}}/{n_{Fe}}=  7/5`

`n_{Mg} = 7x(mol) → n_{Fe}= 5x(mol)`

`Fe_3O_4 = FeO . Fe_2O_3`

Vì `n_{FeO} = n_{Fe_2O_3}` 

Quy `B` về `Fe_3O_4(amol)`

Theo pt: 

`n_{FeCl_3} = 2n_{Fe_3O_4} = 2a(mol)`

Đặt `V_{H_2} = 22,4(l)` 

`n_{H_2} = {22,4}/{22,4} = 1(mol)` 

`n_O= 4n_{Fe_3O_4} = 4a(mol)` 

Bảo toàn `H`: 

`n_{HCl} = 2n_O + 2n_{H_2}`

`= 8a + 2(mol)`

Bảo toàn `Mg`: 

`n_{MgCl_2} = n_{Mg} = 7x(mol)`

Bảo toàn `Fe`: 

`n_{FeCl_2} = n_{Fe} + 3n_{Fe_3O_4}`

`= 5x + 3a(mol)`

Bảo toàn `Cl`: 

`n_{Cl} = 2n_{FeCl_2} + 2n_{MgCl_2}`

`= 2. (5x + 3a) + 2.  7x = 24x + 6a(mol)`

`n_{H^{+}} = n_{Cl^{-}}`

`<=> 8a + 2 = 24x + 6a`

`<=> 24x - 2a = 2(1)`

Bảo toàn `Fe`: 

`n_{Fe_2O_3} = 1/2n_{FeCl_2} `

`= 1/2(5x + 3a) = 2,5x + 1,5a(mol)`

`V` lít `H_2` khử vừa đủ hoàn toàn chất rắn `D` khi đun nóng:

`n_{H_2} = 3n_{Fe_2O_3}`

`<=> 1 = 3(2,5x + 1,5a)(2)`

`(1)(2)→ x= {11}/{123}(mol); a= 3/{41}(mol)`

`m_{Mg} = {11}/{123} .7.   24= {616}/{41}(g)`

`m_{Fe}= {11}/{123}. 5.56= {3080}/{123}(g)`

`m_{A+ B} = {3080}/{123} + {616}/{41} + 3/{41} .232`

`= {7016}/{123}(g)`

`%m_{Mg} = {616}/{41} : {7016}/{123} .100= 26,34%`

`%m_{Fe} = {3080}/{123} : {7016}/{123} .100= 43,9%`