Limiting Reagents

Question

83.0 g of Al reacts with 67.0 g of O2 to produce Al2O3. How much Al2O3 is actually produced assuming 1000% yield?

Answer 1

First write a correctly balanced equation:

4Al(s) + 3O₂(g) ==> 2Al₂O₃(s)

moles Al(s) present = 83.0 g x 1 mol Al/22.99 g = 3.61 moles Al

moles O₂(g) present = 67.0 g x 1 mol O₂/32 g = 2.09 moles O₂

Which reactant is limiting? One of the easiest way to determine limiting reactant is to divide the moles present by the coefficient in the balanced equation. Thus....

For Al, we have 3.61/4 = 0.9 and for O₂ we have 2.09/3 = 0.7 so O₂ is limiting.

The theoretical yield (100%, not 1000%) will be governed by the O₂ since it is limiting:

Theoretical yield of Al₂O₃ = 2.09 mol O2 x 2 mol Al₂O₃ / 3 mol O₂ = 1.39 moles Al₂O₃

Theoretical mass = 1.39 moles Al₂O₃ x 101.96 g/mol = 142 g Al₂O₃