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Stoichiometry Mole–Mole Stoichiometry

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27 g Al reacts with O₂ according to: 4Al + 3O₂ → 2Al₂O₃ How many moles of O₂ are required?

Verified Answer & Explanation⌄

Correct Answer: C) 0.75 mol

Concept Explanation

27 g Al = 1 mol. 4 mol Al require 3 mol O₂. Therefore, 1 mol Al requires 3/4 = 0.75 mol O₂.

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