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Title: You heat 3.886 g of a mixture of Fe3O4 and FeO to form 4.095 g Fe2O3. The mass percent of FeO ... Post by: Dianajupiter1 on Mar 21, 2021 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. Title: You heat 3.886 g of a mixture of Fe3O4 and FeO to form 4.095 g Fe2O3. The mass percent of FeO ... Post by: college98 on Mar 21, 2021 Content hidden
Title: Re: You heat 3.886 g of a mixture of Fe3O4 and FeO to form 4.095 g Fe2O3. The mass percent of FeO ... Post by: acaruso on Jul 11, 2022 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 % |