priscilla89 said:(2 mol Al2O3 / 4 mol Al)
p21bass said:Two things:
1: Don't forget to verify which reagent (Al or O2) is the limiting reagent.
2: Your conversion factors are wrong. For instance - there aren't 108g of Al in 1 mol of Al.
The formula for calculating percent yield of Al2O3 is: (actual yield of Al2O3 / theoretical yield of Al2O3) * 100%.
The actual yield of Al2O3 can be determined by conducting the chemical reaction in a controlled environment and measuring the amount of Al2O3 produced. This can be done by using techniques such as weighing the product or using titration methods.
The theoretical yield of Al2O3 is the maximum amount of Al2O3 that can be produced from a given amount of reactants, assuming all reactants are completely converted into the desired product.
The percent yield of Al2O3 can be affected by various factors such as impurities in the reactants, incomplete reactions, side reactions, and experimental errors.
To improve the percent yield of Al2O3, one can ensure the purity of the reactants, optimize reaction conditions, and carefully follow the experimental procedure. Additionally, conducting multiple trials and using techniques to minimize experimental errors can also improve the percent yield.