Mole Concept and Stoichiometry: Mixture Stoichiometry MCQ

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Mixture StoichiometryHardQuestion 600069

Question

A mixture of FeO and Fe₂O₃ contains 72% Fe by mass. Mole ratio FeO:Fe₂O₃ is:

A
1:1
Correct
B
2:1
C
1:2
D
3:1

Correct answer

1:1

Explanation

Let x mol FeO and y mol Fe₂O₃. Fe mass: 56x + 112y. Total mass: 72x + 160y. 56x+112y = 0.72(72x+160y) → 56x+112y = 51.84x+115.2y → 4.16x = 3.2y → x/y = 3.2/4.16 ≈ 0.77 ≈ not 1:1. Actually: x/y = 3.2/4.16 = 10/13 ≈ 0.77. Closest standard answer: recheck with 70% Fe: if 70%: 56x+112y=0.7(72x+160y) → 56x+112y=50.4x+112y → 5.6x=0 → x=0. At 72%: x/y≈0.77. None match exactly; the closest textbook value for Fe₃O₄ (equal FeO:Fe₂O₃) is 72.4% → approximately 1:1.

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