10.8 g of aluminium reacts with excess oxygen to form aluminium oxide: 4Al + 3O2 → 2Al2O3. What is the theoretical yield of aluminium oxide? (Relative masses: Al = 27.0, Al2O3 = 102.0)

20.410 grams of Al2O3 are produced

Explanation:

First, we need to calculate the number of moles of aluminum that are actually reacting..

     We will convert from grams to moles using the atomic weight which (for aluminum) is 26.982 g/mol (which can also be rewritten as 1 mol/26.982)

10.8 g Al * 1 mol Al/26.982 g Al = 0.4002668446 moles Al

**Grams cancel out, so we are left with moles

(I would type this differently so that it would make more sense visually, however is giving me errors, so try writing it out yourself if it doesn't initially make sense..put the equation in fraction form)

     Now that we have the moles of Al, we will convert from moles of Al reacting to moles of aluminum oxide produced..

To do this we will use the balanced chemical equation..

4Al + 3O2 -> 2Al2O3

Using the balanced chemical equation, we can see that for every 4 moles of aluminum that react, 2 moles of aluminum oxide are produced..

0.4002668446 moles Al * 2 mol Al2O3 / 4 mol Al

= 0.2001334223 moles Al2O3..

Now we need to convert from moles of Al2O3 to grams..

We can do this using the atomic weight of Al2O3..

The atomic weight of Al2O3 is 101.961 g/mol or 1 mol/101.961 grams

Now we will convert to grams of Al2O3...

0.2001334223 moles Al2O3 * 101.961 grams Al2O3/1 mol Al2O3

= 20.40580387..

20.410 grams of Al2O3 are produced