a) Zn +2HCl---->ZnCl2 +H2
0,22---------------------------0,22
m\(_{H2}=0,44\left(g\right)\)
b) Ta có
n\(_{Zn}=\frac{14}{65}=0,22\left(mol\right)\)
m\(_{HCl}=\frac{100.20}{100}=20\left(g\right)\)
n\(_{HCl}=\frac{20}{36,5}=0,55\left(mol\right)\)
=> HCl dư
c) Theo pthh
n\(_{HCl}=2n_{Zn}=0,44\left(mol\right)\)
n\(_{HCl}dư=0,55-0,44=0,11\left(mol\right)\)
C%HCl=\(\frac{0,11.36,5}{100+14-0,22}.100\%=3,53\%\)
Theo pthh
n\(_{ZnCl2}=n_{Zn}=0,22\left(mol\right)\)
C%ZnCl2 =\(\frac{0,22.136}{100+14-0,22}.100\%=26,3\%\)