An ore contains Fe3O4 and no other iron. The
iron in a 54.6-gram sample of the ore is all converted by a series of chemical reactions to
Fe2O3. The mass of Fe2O3 is measured to be
21.3 g. What was the mass of Fe3O4 in the
sample of ore?
Answer in units of g
21.3 g Fe2O3 x (2*molar mass Fe3O4)/(3*molar mass Fe2O3)= ?
30.88g
To find the mass of Fe3O4 in the sample of ore, we need to use the given information about the mass of Fe2O3 and the stoichiometry of the chemical reaction.
Let's start by writing the balanced chemical equation for the conversion of Fe3O4 to Fe2O3:
Fe3O4 → 2Fe2O3
From the balanced equation, we can see that 1 mole of Fe3O4 is equivalent to 2 moles of Fe2O3. We need to find the mass of Fe3O4, so we'll use the molar masses of Fe3O4 and Fe2O3 to convert the mass of Fe2O3 to moles, and then to mass of Fe3O4.
1 mole of Fe3O4 = 3 × 55.845 g/mol + 4 × 16.00 g/mol = 231.54 g/mol
2 moles of Fe2O3 = 2 × (2 × 55.845 g/mol + 3 × 16.00 g/mol) = 319.68 g/mol
Now, we can set up a proportion to find the mass of Fe3O4:
(21.3 g Fe2O3) / (319.68 g/mol Fe2O3) = (mass of Fe3O4) / (231.54 g/mol Fe3O4)
Solving for the mass of Fe3O4:
mass of Fe3O4 = (21.3 g Fe2O3) × (231.54 g/mol Fe3O4) / (319.68 g/mol Fe2O3)
mass of Fe3O4 = 15.35 g
Therefore, the mass of Fe3O4 in the sample of ore is 15.35 g.