PTHH: Zn + 2HCl -----> ZnCl2 + H2
a) nZn=\(\frac{19,5}{65}=0,3\left(mol\right)\)
=>nH2=\(\frac{0,3.1}{1}=0,3\left(mol\right)\)
=> VH2=0,3.22,4=6,72(l)
b) PTHH: Fe2O3 + 3H2 ----------> 2Fe + 3H2O
Ta có : nFe2O3=\(\frac{19,2}{56}=0,34\left(mol\right)\)
Ta có tỉ lệ:
\(\frac{nFe2O3}{2}\)=\(\frac{0,34}{2}\) > \(\frac{nH2}{3}=\frac{0,3}{3}\)
=> Fe2O3 dư => Sử dụng nH2
=> nFe=\(\frac{0,3.2}{3}=0,2\left(mol\right)\)
=> mFe=0,2.56=11,2 (g)