Mình cần gấp ạ , mình cảm ơn trước :**

`n_{NaOH}=0,5.0,5=0,25(mol)`

`m_{Cu}=3,2(g)`

Nếu `FeCl_{2y //x}=FeCl_3`

TN2:

Bảo toàn Cl: `n_{AgCl}=3n_{FeCl_3}+2n_{CuCl_2}={43,975}/{143,5}\approx 0,3(mol)`

TN1:

Theo PT: `n_{NaOH}=3n_{FeCl_3}+2n_{CuCl_2}=0,25(mol)`

Do `0,3\ne 0,25` nên loại.

`->FeCl_{2y //x}=FeCl_2`

`->Fe_xO_y` là `FeO`

`a)`

`FeO+2HCl->FeCl_2+H_2O`

`CuO+2HCl->CuCl_2+H_2O`

P1:

`FeCl_2+2NaOH->Fe(OH)_2+2NaCl`

`CuCl_2+2NaOH->Cu(OH)_2+2NaCl`

P2:

`FeCl_2+2AgNO_3->Fe(NO_3)_2+2AgCl`

`Fe(NO_3)_2+AgNO_3->Fe(NO_3)_3+Ag`

`CuCl_2+2AgNO_3->Cu(NO_3)_2+2AgCl`

`b)`

Mỗi phần đặt `n_{FeCl_2}=x(mol);n_{CuCl_2}=y(mol)`

Theo PT: `n_{AgCl}=n_{NaOH}=2x+2y=0,25(1)`

Lại có `n_{Ag}=n_{Fe(NO_3)_2}=n_{FeCl_2}=x(mol)`

`->0,25.143,5+108x=43,975`

`->x=0,075(mol)->n_{FeO(1P)}=x=0,075(mol)`

`(1)->y=0,05(mol)->n_{CuO(1P)}=y=0,05(mol)`

`n_{Cu(bd)}={3,2}/{64}=0,05(mol)`

Do chia thành 2 phần bằng nhau.

`->` Trong \(Y:\ \begin{cases} n_{Cu}=n_{Cu(bd)}+n_{CuO}=0,05+0,05.2=0,15(mol)\\ n_{Fe}=n_{FeO}=0,075.2=0,15(mol)\\ n_O=n_{FeO}+n_{CuO}=0,075.2+0,05.2=0,25 \end{cases}\)

`m=0,075.2.72+0,05.2.64+3,2=20,4(g)`