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

a. What is the theoretical yield of aluminum oxide if 1.60 mol of aluminum metal is exposed to 1.50 mol of oxygen?

Answer 1

Balanced chemical equation is:
4 Al + 3 O2 ---> 2 Al2O3


4 mol of Al reacts with 3 mol of O2
for 1.6 mol of Al, 1.2 mol of O2 is required
But we have 1.5 mol of O2

so, Al is limiting reagent
we will use Al in further calculation


Molar mass of Al2O3,
MM = 2*MM(Al) + 3*MM(O)
= 2*26.98 + 3*16.0
= 101.96 g/mol

According to balanced equation
mol of Al2O3 formed = (2/4)* moles of Al
= (2/4)*1.6
= 0.8 mol


use:
mass of Al2O3 = number of mol * molar mass
= 0.8*1.02*10^2
= 81.57 g
Answer: 81.6 g