Answer :
Answer: 24.87g Al2O3
Explanation: Aluminum is our limiting reagent
[tex]13.2g Al * \frac{1 mol Al}{26.98 g} * \frac{2 mol Al2O3}{4 mol Al} = 0.244 mol Al2O3[/tex]
First of all we need to convert the grams of aluminium to moles, then use the molar fraction of the balanced equation (4 moles of aluminium equals 2 of aluminum oxide).
If we made the same procedure with the oxigen we get a 0.273 mol of Al2O3, therefore the O2 is the excess reagent.
The last step is convert the moles of the limiting reagent to grams
[tex]0.244 mol Al_{2} O_{3} * \frac{101.96g}{1mol Al2O3} = 24.87g Al2O3[/tex]
And that´s it!