\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(n_{HCl}=\dfrac{12}{36,5}=\dfrac{24}{73}\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{\dfrac{24}{73}}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)