Question

Aluminum oxide is synthesized according to the following reaction: 4Al + 3O2 → 2Al2O3 If 60.72 g of aluminum oxide was formed from 40.81 g oxygen and 50.37 g of aluminum. What is the limiting reactant? What is the theoretical yield? What is the percent yield? Limiting reactant = _____________ Theoretical yield= ______________ Percent yield= _________________

Answer 1

4Al+ 3O2 2 Al₂O₃ . from the stoichiometry, we can observe 4 moles of Al react with 3 moles of O₂ to yield 2 moles of Al₂O₃ . Given 1.275 O₂ moles and 1.866 Al moles combine to form 0.595 moles Al₂O₃ . By the unitary method, 1.866moles of Al require 1.399 moles of O₂ . Hence, provided O₂ moles are not sufficient to complete the whole reaction according to stoichiometry. Therefore, O₂ is the limiting reactant. As O₂ is the limiting reactant in the reaction therefore all the moles of O₂ will get over. Hence, by the unitary method 0.85 moles of Al₂O₃ will form or 86.66gm Al₂O₃ . Dividing actual yield = 60.72 gm by theoretical yield= 86.66gm we get percent yield= 70.066 percent.

Limiting reactant = O₂

Theoretical yield=86.66gm

Percent yield=  70.066%