4Al + 3O2 = 2Al2O3. If 41.5 g aluminum and 69.0 g oxygen undergo a reaction that has a 67.0% yield, What mass of aluminum oxide forms. Please help by showing steps to find answers. Thanks

Answer 1

This is a limiting reagent problem. You know that because amounts for BOTH reactants are given. I solve these problems by working TWO stoichiometry problems. Here is a worked example for that. Print this and save it. It will solve 99.9% of the stoichiometry problems you will have.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Use this method to solve for moles Al2O3 formed with 41.5 g Al. Work a second problem for Al3O3 formed with 69.0 g oxygen. Two answers but in limiting reagent problems the smaller value is ALWAYS the correct one and the reagent producing that value is the limiting reagent. The moles you find will be the theoretical yield (;i.e., 100%). Convert to grams and multiply by 0.67 since it is only 67% efficient.