Em tham khảo nha:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1\,mol\\
{n_{HCl}} = 0,2 \times 2 = 0,4\,mol\\
{n_{Zn}} < \dfrac{{{n_{HCl}}}}{2} \Rightarrow HCl \text{ dư } \\
{n_{HCl}} \text{ dư } = 0,4 - 0,1 \times 2 = 0,2\,mol\\
c)\\
{n_{{H_2}}} = {n_{Zn}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,1\,mol\\
{m_{ZnC{l_2}}} = 0,1 \times 136 = 13,6g
\end{array}\)

 

`n_(Zn)=(6,5)/65=0,1(mol)`

Đổi $200(ml)=0,2(l)$

`=>n_(HCl)=0,2.2=0,4(mol)`

$a)PTHH :$

$Zn+2HCl \to ZnCl_2+H_2$

$b)$

Xét tỉ lệ :

`(n_(Zn))/1=0,1<0,2=(n_(HCl))/2`

`=>Zn` hết, $HCl$ dư

`=>Mol` tính theo $Zn$ 

Theo `PT,n_(HCl)(pứ)=2n_(Zn)=0,2(mol)`

`=>n_(HCl)(dư)=0,4-0,2=0,2(mol)`

$c)$

Theo $PT:$

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

`=>V_(H_2)(đktc)=0,1.22,4=2,24(l)`

`=>m_(ZnCl_2)=0,1.136=13,6(g)`