Question

When exposed to air, aluminum metal, Al, reacts with oxygen, O2, to produce a protective coating of aluminum oxide, Al2O3, which prevents the aluminum from rusting underneath. The balanced reaction is shown here: 4Al+3O2→2Al2O3

What is the theoretical yield of aluminum oxide if 2.20 mol of aluminum metal is exposed to 1.95 mol of oxygen?

Answer 1

Given balanced equation is 4Al+3O₂ → 2Al₂O₃

According to the balanced equation,

4 moles of Al reacts with 3 moles of O₂       

2.20 moles of Al reacts with M moles of O₂       

M = (2.20x3)/4

   = 1.65 moles

So 1.95-1.65 = 0.3 moles of Oxygen left unreacted.It is the excess reactant

Since all the mass of Al completly reacted it is the limiting reactant.

From the equation ,

4 moles of Al produces 2 moles Al₂O₃

2.20 moles of Al produces N moles Al₂O₃

N = (2.20x2)/4

   = 1.1 moles

Molar mass of Al₂O₃ is (2xAt.mass of Al) + (3xAt.mass of O)

                                  = (2x27) + (3x16)

                                  = 102 g/mol

So mass of Al₂O₃ is , m = number of moles x molar mass

                                    = 1.1 mol x 102 g/mol

                                   = 112.2 g

Therefore the theoretical yield of aluminum oxide is 112.2 g