A mixture of FeO and Fe2O3 with a mass of 10.0g is converted to 7.43g of pure Fe. What are the amounts in grams of FeO and Fe2O3?

Molar mass of Fe = 55.85 g/mol
Molar mass of FeO = (55.85 + 16.00) g/mol = 71.85
Molar mass of Fe₂O₃ = (55.85×2 + 16.00×3) g/mol = 159.70 g/mol

Mass fraction of Fe in FeO = 55.85/71.85
Mass fraction of Fe in Fe₂O₃ = 55.85 × 2 / 159.70 = 111.70/159.70

Let x g be the mass of FeO in the mixture.
Then, the mass of Fe₂O₃ in the mixture = (10 - x) g

(Mass of Fe in FeO) + (Mass of Fe in Fe₂O3) = 7.43 g
(55.85/71.85)x + (1117.0/159.70) - (111.70/159.70)x = 7.43
[(55.85/71.85) - (111.70/159.70)]x = 7.43 - (1117.0/159.70)
x = [7.43 - (1117.0/159.70)] / [(55.85/71.85) - (111.70/159.70)]
x = 5.59
10 - x = 4.41

Amount of FeO in the mixture = 5.59 g
Amount of Fe₂O₃ in the mixture = 4.41 g