nZn=0,1(mol)
PTHH: Zn + 2 HCl -> ZnCl2 + H2
nH2=nZn=0,1(mol)
=>V(H2,đktc)=0,1.22,4=2,24(l)
=> CHỌN A
\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.1................................0.1\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(\Rightarrow A\)