High School

An ore contains Fe\(_3\)O\(_4\) 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 Fe\(_2\)O\(_3\). The mass of Fe\(_2\)O\(_3\) is measured to be 29 g. What was the mass of Fe\(_3\)O\(_4\) in the sample of ore?

Answer :

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.

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