Giúp mình câu 29 vs ạ

`a)`

Đặt `n_{Fe}=x(mol);n_{Zn}=y(mol)`

`->56x+65y=4,28(1)`

`Fe+2HCl->FeCl_2+H_2`

`Zn+2HCl->ZnCl_2+H_2`

Theo PT: `n_{FeCl_2}=x(mol);n_{ZnCl_2}=y(mol)`

`->127x+136y=9,25(2)`

`(1)(2)->x=0,03;y=0,04`

`\%m_{Fe}={0,03.56}/{4,28}.100\%=39,25\%`

`\%m_{Zn}=100-39,25=60,75\%`

`b)`

`n_{HCl\ pu}=2x+2y=0,14(mol)`

`n_{HCl\ bd}=0,5.0,4=0,2(mol)`

`->n_{HCl\ du}=0,2-0,14=0,06(mol)`

`n_{NaOH}=n_{Ba(OH)_2}=0,02V(mol)`

`NaOH+HCl->NaCl+H_2O`

`Ba(OH)_2+2HCl->BaCl_2+2H_2O`

`->n_{HCl\ du}=n_{NaOH}+2n_{Ba(OH)_2}`

`->0,06=0,02V+0,04V->V=1`

Vậy cần 1L dung dịch B

`c)`

A chứa `FeCl_2(0,03);ZnCl_2(0,04);HCl(0,06)`

`HCl+NaOH->NaCl+H_2O`

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

`ZnCl_2+2NaOH->Zn(OH)_2+2NaCl`

`->n_{NaOH}=2n_{FeCl_2}+n_{HCl}+2n_{ZnCl_2}=0,2(mol)`

`->V_{dd\ NaOH}={0,2}/1=0,2(L)`