a) Zn +2HCl--->ZnCl2 +H2
n\(_{Zn}=0,1\left(mol\right)\)
n\(_{HCl}=1\left(mol\right)\)
=> HCl dư
Theo pthh
n\(_{H2}=n_{Zn}=0,1\left(mol\right)\)
V\(_{H2}=0,1.22,4=2,24\left(l\right)\)
b) n HCl dư=0,8(mol)
CM(HCl dư) =\(\frac{0,8}{0,2}=4M\)
n ZnCl2 =0,1
=> CM(ZnCl2)=\(\frac{0,1}{0,2}=0,5\left(M\right)\)