\(n_{Zn}=\dfrac{6,5}{65}=0,1(mol)\\ PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ \Rightarrow n_{ZnSO_4}=n_{Zn}=0,1(mol)\\ \Rightarrow m_{ZnSO_4}=0,1.161=16,1(g)\)
Bảo toàn nguyên tố Zn:
\(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow m_{ZnSO_4}=16,1\left(g\right)\)