$$4Al + 3O_2 \rightarrow 2Al_2O_3$$
From the stoichiometry of the balanced equation, we see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. Therefore, the limiting reactant in this case is oxygen.
To calculate the theoretical yield of aluminum oxide, we need to determine the number of moles of oxygen available for the reaction. We are given that there are 2.55 mol of oxygen. Using the stoichiometry of the balanced equation, we can calculate the maximum number of moles of aluminum oxide that can be produced:
$$2.55 \text{ mol }O_2 \times \frac{2 \text{ mol }Al_2O_3}{3 \text{ mol }O_2} = 1.70 \text{ mol }Al_2O_3$$
Therefore, the theoretical yield of aluminum oxide is 1.70 mol.
