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Question

You heat 3.854 g of a mixture of Fe3O4 and FeO to form 4.148 g Fe2O3. The mass percent of FeO originally in the mixture was: A) 92.9% B) 55.8% C) 44.2% D) 38.5% E) none of these

Answer 1

Molar masses:

Fe₂O₃ = 159.69 g/mol

FeO = 71.845 g/mol

Fe₃O₄ = 231.533 g/mol

4.148 g Fe₂O₃ x 1 mol Fe₂O₃/159.69 g x 2 mol Fe/mol Fe₂O₃ x 55.85 g Fe/mol Fe = 2.901 g Fe in the product

Let x = g FeO and let y = g Fe₃O₄

x + y = 3.854 g

x = 3.854 g - y

2.901 g Fe = (3.854 - y)(55.85 g Fe/71.845 g FeO/mol) + y(167.55 g Fe/231.533 g Fe₃O₄/mol)

y = 1.679 g Fe₂O₃

x = 3.854 g - 1.679 g = 2.157 g FeO

Mass % FeO = 2.157 g/3.854 g (x100%) = 55.9% (ANSWER B)