what is the right formula?

Question

Suppose synthesis of Al2O3 is done in the lab using 20.50 g of Al and excess of O2. Upon purification, 35.48 grams of Al2O3 is produced. What is the percent yield of Al2O3?

Answer 1

First, write a correctly balanced equation for this reaction: 4Al(s) + 3O₂(g) ==> 2Al₂O₃(s)

From this, note that 4 moles Al react with 3 moles O₂ to produce 2 moles Al₂O₃.

Moles of Al used = 20.50 g Al x 1 mole Al/26.98 g = 0.7598 moles Al

Since there is excess O₂, Al will be limiting and will determine moles of Al₂O₃ formed.

Moles Al₂O₃ formed = 0.7598 moles Al x 2 moles Al₂O₃/4 moles Al = 0.3799 moles Al₂O₃ formed theoretically.

Grams Al₂O3 theoretically formed = 0.3799 moles x 101.96 g/mole = 38.74 g

% yield = 35.48 g/38.74 g (x100%) = 91.58%