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

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

`a)`

Theo phương trình : `n_{ZnCl_2} = n_{Zn} = 0,01` `mol`

`=> m_{ZnCl_2} = 0,01 . 136 = 1,36` `g`

`b)`

Theo phương trình : `n_{H_2} = n_{Zn} = 0,01` `mol`

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

`c)`

Theo phương trình : `n_{HCl} = 2n_{Zn} = 0,02` `mol`

`=> V_{\text{dung dịch HCl}} = (0,02)/(1) = 0,02` `l`

$a)$

$n_{Zn}=\dfrac{0.65}{65}=0.01(mol)$

$Zn+2HCl \to ZnCl_2+H_2$

$0.01$`->`$0.02$`->`$0.01$`->`$0.01$ $(mol)$

$n_{ZnCl_2}=0.01(mol)$

`=>`$mn_{ZnCl_2}=0.01 \times (65+35.5 \times 2)=1.36(gam)$

$b)$

$n_{H_2}=0.01(mol)$

`=>`$V_{H_2}=0.01 \times 24.79=0.2479(L)$

$c)$

$n_{HCl}=0.02(mol)$

$V=\dfrac{n}{C_M}=\dfrac{0.02}{1}=0.02(L)$