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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Dianajupiter1 Dianajupiter1
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3 years ago
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

Chemistry: An Atoms First Approach


Edition: 3rd
Authors:
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college98college98
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#2
A year ago
Fe2O3 moles = mass / Molar mass of Fe2O3

= 4.095 g / 159.7 g/mol

= 0.02564

Let FeO moles = m , Fe3O4 moles = 0.02564-m

Mass of FeO = moles x molar mass of FeO

= m x 71.844

mass of Fe3O4 = (0.02564 -m) x 231.533 g/mol

= 5.9365 - 231.533m

total mass of FeO + Fe3O4 = 3.886

71.844 m + 5.9365 - 231.533 m = 3.886

m = 0.01284

FeO mass = 71.844 g/mol x 0.01284 mol = 0.9225g

mass % of FeO = ( 100 x FeO mass/ sample mass)

= ( 100 x 0.9225 g / 3.886 g)

= 23.7 %
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