`n_{Zn}=(65)/(65)=1(mol)`

`n_{HCl}=(200.14,6\%)/(36,5)=0,8(mol)`

PTHH:

`Zn+2HCl->ZnCl_2+H_2`

Lập tỉ lệ :

`(n_{Zn})/(1) =(1)/(1) > (n_{HCl})/2 =(0,8)/2`

`=>` Sau phản ứng `Zn` dư, `HCl` hết

`=>n_{Zn pứ}= n_{H_2}=n_{ZnCl_2}=1/2 n_{HCl}= 1/2 .0,8=0,4(mol)`

`=>m_{dd sau pứ}=m_{Zn pứ}+m_{dd HCl}-m_{H_2}`

`=0,4.65+200-0,4.2=225,2(g)`

`C\%_{ZnCl_2}=(0,4.136)/(225,2).100\%≈24,16\%`

n$_{zn}$=m= $\frac{m}{M}$= $\frac{65}{65}$ = 1 mol

m$_{HCL}$= $\frac{m_{dd} . C %}{ 100}$ = $\frac{200.14,6}{100}$ = 29,2

n$_{HCL}$ = $\frac{m}{M}$ = $\frac{22,9}{36.5}$ =0.8 mol

PTHH Zn+2HCL ->ZnCl $_{2}$ + H$_{2}$ O

1      0.8 => zn dư

Zn+2HCL ->ZnCl $_{2}$ + H$_{2}$ O

0.4---0.8----0.4-------0.4 (mol)

m dd sau phản ứng= 200+65-0,4.2=264.2(g)

%C zncl2= $\frac{0,4.136}{264.2}$ . 100%=20,59%

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