\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(n_{HCl}=0.2\cdot2=0.4\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có :
\(\dfrac{0.1}{1}< \dfrac{0.4}{2}\rightarrow HCldư\)
\(n_{H_2}=0.1\left(mol\right)\)
\(V=0.1\cdot22.4=2.24\left(l\right)\)