làm giúp em bài 8 vs ạ

8.

`a)`

`Zn+2HCl->ZnCl_2+H_2`

`Mg+2HCl->MgCl_2+H_2`

Đặt `n_{Zn}=x(mol);n_{Mg}=y(mol)`

`->65x+24y=17,8(1)`

Theo PT: `n_{H_2}=x+y={8,96}/{22,4}=0,4(2)`

`(1)(2)->x=y=0,2(mol)`

`m_{Zn}=0,2.65=13(g)`

`m_{Mg}=0,2.24=4,8(g)`

`b)`

`n_{HCl}={182,5.20\%}/{36,5}=1(mol)`

Theo PT:

`n_{HCl\ pu}=2n_{H_2}=0,8(mol)`

`n_{MgCl_2}=y=0,2(mol)`

`n_{ZnCl_2}=y=0,2(mol)`

`m_{dd\ spu}=17,8+182,5-0,4.2=199,5(g)`

`->C\%_{MgCl_2}={0,2.95}/{199,5}.100\%\approx 9,52\%`

        `C\%_{ZnCl_2}={0,2.136}/{199,5}.100\%\approx 13,63\%`

        `C\%_{HCl\ du}={(1-0,8).36,5}/{199,5}.100\%\approx 3,66\%`

`c)`

`-` Vừa đủ trung hòa:

`NaOH+HCl->NaCl+H_2O`

`->n_{NaOH}=n_{HCl\ du}=1-0,8=0,2(mol)`

`->V_{dd\ NaOH}={0,2}/2=0,1(L)=100(mL)`

`-` Tủa cực đại:

`NaOH+HCl->NaCl+H_2O`

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

`2NaOH+MgCl_2->Mg(OH)_2+2NaCl`

`->n_{NaOH}=n_{HCl}+2n_{ZnCl_2}+2n_{MgCl_2}=1(mol)`

`->V_{dd\ NaOH}=1/2=0,5(L)=500(mL)`

`-` Tủa cực tiểu: 

`NaOH+HCl->NaCl+H_2O`

`4NaOH+ZnCl_2->Na_2ZnO_2+2NaCl+2H_2O`

`2NaOH+MgCl_2->Mg(OH)_2+2NaCl`

`->n_{NaOH}=n_{HCl}+4n_{ZnCl_2}+2n_{MgCl_2}=1,4(mol)`

`->V_{dd\ NaOH}={1,4}/2=0,7(L)=700(mL)`