# Chapter 4 - Sections 4.1-4.9 - Exercises - Problems by Topic: 38b

Aluminum is the excess reactant, and therefore $O_{2}$ is the limiting reactant.

#### Work Step by Step

Determine the limiting reactant. Choose one of the reactants and determine how much we need compared to how much is available. We choose to compare Al. $4$ $mol$ $Al (avaliable)$ $2.6 mol$ $O_{2}$$\times\frac{4 mol Al}{3 mol O_{2}}$ $=3.5$ $mol$ $Al (needed)$

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