20.0g sample of Al reacts to produce 32.7g of Al2O3, what is the percentage yield of the reaction

Question

reaction; 4al(s) + 30,2(g) ----- 2Al,2O,3

Answer 1

4Al(s) + 3O₂(g) ==> 2Al₂O₃(s) ... balanced equation

Assuming O₂ is not limiting, we have the following:

Theoretical yield of Al₂O₃ = 20.0 g Al x 1 mol Al / 27.0 g x 2 mol Al₂O₃ / 4 mol Al x 102 g/mol = 37.8 g Al₂O₃

% yield = actual yield / theoretical yield (x100%)

32.7 g / 37.8 g (x100%) = 86.5% yield