mol HCl =0,5*0,2=0,1mol
=> HCl hết , Zn dư
pthh Zn + 2HCl --> ZnCl2 + H2
0,1 0,05
=> vH2 =0,05*22,4=1,12l
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{HCl}=0,2\cdot0,5=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,5}{1}>\dfrac{0,1}{2}\) \(\Rightarrow\) Kẽm còn dư, HCl p/ứ hết
\(\Rightarrow n_{H_2}=0,05\left(mol\right)\) \(\Rightarrow V_{H_2}=0,05\cdot22,4=1,12\left(l\right)\)