1, mCuO = 80% .30 = 12(g)
=>nCuO=12/80=0,15(mol)
mFeO = 30-12=18(g)
=> nFeO = 18/72=0,25(mol)
Ta có PT
CuO + H2 ---> Cu + H2O
0,15..0,15.........0,15..0,15
FeO + H2 ---> Fe + H2O
0,25..0,25......0,25...0,25
mCu = 0,15.64=9,6(g)
mFe = 0,25.56=14(g)
nH2= 0,15+ 0,25 = 0,4(mol)
=> VH2= 0,4.22,4 = 8,96(l)
2, nO(Oxit)=12nHCl=12.0,15.3=0,225mol
nFe=8,456=0,15mol
-Gọi công thức là FexOy
-Ta có:
x:y=0,15:0,225=2:3→
Fe2O3