`a,`

`PTHH:Zn+2HCl->ZnCl_2+H_2`

`b,`

Đổi: `200ml=0,2l`

Theo bài ra: `n_{Zn}=(9,75)/(65)=0,15(mol)`

`n_{HCl}=0,2.1=0,2(mol)`

Theo PTHH, xét tỉ lệ:

`n_{Zn}/1=(0,15)/1>n_{HCl}/2=(0,2)/2`

`=>` Sau phản ứng `Zn` còn dư, `HCl` hết. Tính theo `HCl`

Theo PTHH: `n_{ZnCl_2}=n_{H_2}=0,5n_{HCl}=0,5.0,2=0,1(mol)`

`=>V_{H_2}=0,1.24,79=2,479(l)`

`c,`

`C_{MddZnCl_2}=(0,1)/(0,2)=0,5M`

`a)`

`n_{Zn}={9,75}/{65}=0,15(mol)`

`n_{HCl}=0,2.1=0,2(mol)`

`Zn+2HCl->ZnCl_2+H_2`

`0,1←0,2->0,1->0,1(mol)`

`0,15>{0,2}/2->Zn` dư

`b)`

`V_{H_2}=0,1.24,79=2,479(L)`

`c)`

`C_{M\ ZnCl_2}={0,1}/{0,2}=0,5M`