Question

An ore contains fe3o4 and no other iron. the iron in a 36.6-gram sample of the ore is all converted by a series of chemical reactions to fe2o3. the mass of fe2o3 is measured to be 29 g. what was the mass of fe3o4 in the sample of ore?

Answer 1

To solve this problem, let us first find for the molar mass of Fe2O3 and Fe3O4.

Fe = 55.85 g/mol and O = 16 g/mol

Therefore,

Fe2O3 = 159.7 g/mol

Fe3O4 = 231.55 g/mol

We are given that there are 29 g of Fe2O3, we calculate for the amount of Fe from this in moles:

mol Fe = 29 g Fe2O3 (1 / 159.7 g/mol) (2 mol Fe / 1 mol Fe2O3)

mol Fe = 0.363 mol

Converting this to Fe3O4:

mass Fe3O4 = 0.363 mol Fe (1 mol Fe3O4 / 3 mol Fe) (231.55 g/mol)

mass Fe3O4 = 28.03 g

Therefore there are 28.03g of Fe3O4 in the ore.