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

Question

Quizplus · Accepted Answer

Firstly, let's determine the amount of FeO reacted:
FeS1U1B13S1U1B0 + FeO -> FeS1U1B12S1U1B0O1U1B1S1U1B0
Molar mass of FeS1U1B13S1U1B0 = 87.92 g/mol
Molar mass of FeO = 71.85 g/mol
Molar mass of FeS1U1B12S1U1B0O1U1B1S1U1B0 = 87.00 g/mol
0.044 moles of FeS1U1B13S1U1B0 is equivalent to 0.044 moles of FeO reacted (according to balanced chemical equation above)
Therefore, the amount of FeO originally in the mixture is:
(0.044 moles / 3.854 g) x (55.85 g/mol) x 100% = 55.8%

You Heat 3