The total mass of aluminum hydroxide at the end of the reaction was 0.94g. What was the percent yield of this experiment?

Question

A student in a chemistry lab mixes 13.8g of aluminum sulfate with 2.04g calcium hydroxide to produce aluminum hydroxide and calcium sulfate.

The unbalanced reaction is: Al₂(SO₄)₃ + Ca(OH)₂  Al(OH)₃+ CaSO₄

Answer 1

Assuming this is an extension of the previous question that you posted, the % yield can be calculated as follows:

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

From your previous question, we calculated the theoretical yield as

1. Theoretical yield of Al(OH)₃ = 0.0275 mols Ca(OH)₂ x 2 mol Al(OH)₃ / 3 mol Ca(OH)₂ x 78.0 g / mol = 1.43 g Al(OH)₃

Thus, % yield = 0.94 g / 1.43 g (x100%) = 66% yield