Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)
The maximum amount of Al₂O₃ that can be produced from 5.0 g of Al and 5.0 g of O₂is ________ g.

Question

Quizplus · Accepted Answer

First, we need to determine which reactant is limiting by using stoichiometry. We can convert both Al and O2 to moles: 5.0 g Al * (1 mol Al/27.0 g Al) = 0.185 mol Al 5.0 g O2 * (1 mol O2/32.0 g O2) = 0.156 mol O2 Then, we can use the balanced equation to determine the amount of Al₂O₃ produced: 4 mol Al : 3 mol O2 : 2 mol Al₂O₃ 0.185 mol Al : 0.156 mol O2 : x mol Al₂O₃ Since O2 is the limiting reactant, we use its amount to determine the amount of Al₂O₃ produced: 0.156 mol O2 * (2 mol Al₂O₃/3 mol O2) * (101.96 g Al₂O₃/1 mol Al₂O₃) = 9.4 g Al₂O₃ Therefore, the maximum amount of Al₂O₃ that can be produced is 9.4 g, which corresponds to choice C.

Solid Aluminum and Gaseous Oxygen React in a Combination Reaction