What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?

71.372 g or 0.7 moles

Explanation:

We are given;

Moles of Aluminium is 1.40 mol Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 * 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 * 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al * 0.5

= 1.4 moles * 0.5

= 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles * 101.96 g/mol

= 71.372 g