What is the theoretical yield of aluminum oxide if 2.20 mol of aluminum metal is exposed to 1.95 mol of oxygen? express your answer with the appropriate units?

4Al + 3O₂ ---> 2Al₂O₃

1) By looking at the balanced equation, you can see that 4 mol of Al needs 3 moles of O₂ to fully react . So to react 2.20 moles, how many moles of O₂  are necessary?
 4 moles of Al 3 O₂
2,20 moles of Alx
x= 1,65 moles of O₂  are needed.

You have more than 1,65 moles. In fact, there are 1,95 moles of O2. This means the Al is the limiting reagent.

2)For the theoretical yield to be maximum amount of product possible we need the limiting reagent, aluminum,  to fully react.

So how many moles of Al₂O₃ we obtain, from the 2,20 moles of Al available to react?

 4 moles of Al2 moles of Al₂O₃
2,20 moles of Al x
x= 1,1 moles of Al₂O₃
the theoretical yield  = 1,1 moles

the molar mass of a compound is the sum of the products of the atomic masses by the number of atoms of the element.
molar mass of Na₂SO₄ is - 142 g/mol.
1 mol of Na₂SO₄ has a mass of 142 g.
In 1 mol of Na₂SO₄ the mass of Na is 23 g/mol x 2 = 46 g.
                             
Mass of Na in 1 mol of Na₂SO₄ is - 46 g
                           
mass of Na in 0.820 mol of Na₂SO₄ - 46 g /1 mol x 0.820 mol = 37.72 g.
mass of Na is 37.72 g