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# You heat 3.886 g of a mixture of Fe3O4 and FeO to form 4.095 g Fe2O3. The mass percent of FeO ...

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You heat 3.886 g of a mixture of Fe3O4 and FeO to form 4.095 g Fe2O3. The mass percent of FeO originally in the mixture was

94.9%.

26.1%.

73.9%.

18.0%.

None of these are correct.
Textbook

## Chemistry: An Atoms First Approach

Edition: 3rd
Authors:
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Answer verified by a subject expert
college98college98
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### Related Topics

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3 weeks ago
 Fe2O3 moles = mass / Molar mass of Fe2O3= 4.095 g / 159.7 g/mol= 0.02564Let FeO moles = m , Fe3O4 moles = 0.02564-mMass of FeO = moles x molar mass of FeO= m x 71.844mass of Fe3O4 = (0.02564 -m) x 231.533 g/mol= 5.9365 - 231.533mtotal mass of FeO + Fe3O4 = 3.88671.844 m + 5.9365 - 231.533 m = 3.886m = 0.01284FeO mass = 71.844 g/mol x 0.01284 mol = 0.9225gmass % of FeO = ( 100 x FeO mass/ sample mass)= ( 100 x 0.9225 g / 3.886 g)= 23.7 %