Cách 1:  

PTHH: `Zn` `+` $H_{2}$$SO_{4}$ $\longrightarrow$ $ZnSO_{4}$ `+` $H_{2}$ 

$n_{Zn}$ `=` $\dfrac{6,5}{65}$ `=` `0,1` `(mol)` 

Theo PT, ta có: $n_{ZnSO4}$ `=` $n_{Zn}$ `=` `0,1` `mol` 

⇒ $m_{ZnSO4}$ `=` `16,1` `(g)` 

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Cách 2: 

PTHH: PTHH: `Zn` `+` $H_{2}$$SO_{4}$ $\longrightarrow$ $ZnSO_{4}$ `+` $H_{2}$  

Áp dụng ĐLBTKL, ta có: $m_{ZnSO4}$ `=` `(6,5+9,8)` `-` `0,2` `=` `16,1` `(g)` 

⇒ $m_{ZnSO4}$ `=` `16,1` `(g)` 

Áp dụng ĐLBTKL, ta có:

         `m_{Zn}``+``m_{H_2SO_4}``=``m_{ZnSO_4}``+``m_{H_2}`

  `⇔``6,5``+``9,8``=``m_{ZnSO_4}``+``0,2`

  `⇔``16,3``=``m_{ZnSO_4}``+``0,2`

  `⇔``m_{ZnSO_4}``=``16,1` `(g)`