a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=1.0,2=0,2\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\) => Zn hết, H2SO4 dư
b)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
0,1--->0,1------->0,1
=> \(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)
c) \(\left\{{}\begin{matrix}C_{M\left(H_2SO_4.dư\right)}=\dfrac{0,2-0,1}{0,2}=0,5M\\C_{M\left(ZnSO_4\right)}=\dfrac{0,1}{0,2}=0,5M\end{matrix}\right.\)