a) PTHH: Zn + 2HCl -> ZnCl2 + H2
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PTHH và đề bài, ta có: \(n_{HCl}=2.n_{Zn}=2.0,1=0,2\left(mol\right)\\ n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
b) \(V_{ddHCl}=\dfrac{n_{HCl}}{C_{MddHCl}}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)
c) \(V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)