`a/` `PTHH` là:

          `Zn``+``2HCl``→``ZnCl_2``+``H_2`

`b/` `nZn``=``\frac{m}{M}``=``\frac{6,5}{65}``=``0,1` `(mol)`

`PTHH`            `Zn``+``2HCl``→``ZnCl_2``+``H_2`

                        `0,1`       `0,2`           `0,1`           `0,1`      `mol`

`→``nHCl``=``nZn``=``\frac{0,1.2}{1}``=``0,2` `(mol)`

`→``mHCl``=``n``.``M``=``0,2``.``36,5``=``7,3` `(g)`

`c/`Theo pt, ta có: `nZnCl_2``=``nZn``=``0,1` `(mol)`

`→``mZnCl_2``=``n``.``M``=``0,1``.``136``=``13,6` `(g)` 

`d/`Theo pt, ta có: `nH_2``=``nZn``=``0,1` `(mol)`

`→``VH_2``=``n``.``24,79``=``0,1``.``24,79``=``2,479` `(l)`

   `flower`

Đáp án + Giải thích các bước giải:

`a.`

`PTHH: Zn+2HCl→ZnCl_2+H_2` 

`n_{Zn}=m_{Zn}/M_{Zn}=(6,5)/65=0,1(mol)`

`-` Theo PT:

`+` `n_{HCl}=2.n_{Zn}=2.0,1=0,2(mol)`

`+` `n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1(mol)`

`b.`

`m_{HCl}=M_{HCl}.n_{HCl}=0,2×36,5=7,3(g)`

`c.` 

`m_{ZnCl_2}=M_{ZnCl_2}×n_{ZnCl_2}=0,1×136=13,6(g)`

`d.`

`V_{H_2}=n_{H_2}×24,79=0,1×24,79=2,479(l)`