How many moles of aluminum oxide ai203 can be produced from 12.8 moles of oxygen gas o2 reacting with excess aluminum al

4Al + 3O₂ > 2Al₂O₃
       3 moles     2 moles



3 moles of O₂ 2 moles of Al₂O₃
12,8 moles of O₂  X
X = (12,8×2)/3 = 25,6/3
X = 8,533 moles of Al₂O₃

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