`n_{Zn} = (0,65)/(65) = 0,01` mol

`Zn + 2HCl -> ZnCl_2 + H_2`

`n_{H_2} = n_{Zn} = 0,01 ` mol

`=> V_{H_2} = 0,01 . 24,79 = 0,2479l`

`50ml = 0,05l`

`n_{ZnCl_2} = n_{Zn} = 0,01` mol

`=> C_{M dd ZnCl_2} = (0,01)/(0,05) = 0,2M`

$n_{Zn}=\dfrac{0,65}{65}=0,01(mol)$ 

$Zn +2HCl \xrightarrow{} ZnCL_2 +H_2$ 

theo pt 

$n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,01(mol)$ 

$V_{H_2(đktc)}=0,01.22,4=0,224 (l)$

$C_{M_{ZnCl_2}}=\dfrac{0,01}{0,05}=0,2M$