A 5.430 g mixture of FeO and Fe
3O
4 is reacted with excess of oxygen to form 5.779 g Fe
2O
3. Find the masses of FeO and Fe
3O
4 present in the mixture.
I found the answer by doing the following:
Fe
2O
3 = 159.69 g/mol
FeO = 71.745 g/mol
Fe
3O
4 = 231.535 g/mol
5.779 g Fe
2O
3 / 159.69 g/mol = 0.03619 mol Fe
2O
3x + y
0.03619 mol Fe
2O
3, where x = FeO and y = Fe
3O
4.
This implies that y = 0.03619 - x.
5.430 g = x*71.745 g/mol + (0.03619 - x)*231.535g/mol
x = 0.01846 mol * 71.745 g/mol = 1.324 g FeO
y = 0.03619 - 0.01846 = 0.01773 mol * 231.535 g/mol = 4.105 g Fe
3O
4My question is why doesn't the following work:4FeO + O
2 2Fe
2O
34Fe
3O
4 + O
2 6Fe
2O
3_______________________
2FeO + 2Fe
3O
4 + O
2 4Fe
2O
35.779 g Fe
2O
3 / 159.69 g/mol = 0.03619 mol Fe
2O
3Product:reactants in question are in a 2:1 ratio as given by the stoichiometric coefficients.
0.03619 mol / 2 = 0.01810 mol of FeO and 0.01810 mol Fe
3O
4.
0.01810 mol FeO * 71.745 g/mol = 1.299 g FeO
0.01810 mol Fe
3O
4 * 231.535 g/mol = 4.191 g Fe
3O
4Obviously this doesn't work because the original mixture is 5.430 g and not 1.299 g + 4.191 g = 5.490 g. Why doesn't this work the way I think it should? Thank you.