Zn + 2HCl -> ZnCl2 + H2 (1)
nZn=0,1(mol)
Từ 1:
nH2=nZn=0,1(mol)
VH2=0,1.22,4=2,24(lít)
Zn + 2HCl → ZnCl2 + H2
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)