a)

`PTHH:Zn+2HCl->ZnCl_2+H_2↑`

b)

Ta có: `n_(Zn)=m/M=(1,2)/65≈0,02(mol)`

Theo `PTHH`,ta thấy: `n_(Zn)=n_(H_2)=0,02(mol)`

`=>V_(H_2)=n.22,4=0,02.22,4=0,448(l)`

c)

Theo `PTHH`,ta thấy: `n_(HCl)=2.n_(Zn)=2.0,02=0,04(mol)`

`=>m_(HCl)=n.M=0,04.36,5=1,46(g)`

`->C%ddHCl=(1,46)/200×100%=0,73%`

d)

Theo Định luật bảo toàn khối lượng:

`m_(Zn)+m_(ddHCl)=m_(ddZnCl_2)+m_(H_2)`

`=>m_(ddZnCl_2)=(m_(Zn)+m_(ddHCl))-m_(H_2)`

`=>m_(ddZnCl_2)=(1,2+200)-(0,02.2)`

`=>m_(ddZnCl_2)=201,16`

Theo `PTHH`,ta thấy: `n_(ZnCl_2)=n_(Zn)=0,02(mol)`

`=>m_(ZnCl_2)=n.M=0,02.136=2,72(g)`

`->C%ddZnCl_2=(2,72)/(201,16)×100%≈1,4%`

Đáp án:

Giải thích các bước giải:

 a) `Zn+2HCl→ZnCl_2+H_2↑`

b) `n_{Zn}=\frac{1,2}{65} \approx 0,02\ mol`

`n_{H_2}=n_{Zn}=0,02\ mol`

`V_{H_2}=0,02.22,4=0,448\ lít`

c) `n_{HCl}=2n_{Zn}=0,04\ mol`

`m_{HCl}=0,04.36,5=1,46\ gam`

`C%_{HCl}=\frac{1,46}{200}.100%=0,73%`

d) `m_{dd}=m_{Zn}+m_{dd\ HCl}-m_{H_2}`

`m_{dd}=1,2+200-(0,02.2)=201,16\ gam`

`C%_{ZnCl_2}=\frac{0,02.136}{201,16}.100% \approx 1,35%`