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