First, write a correctly balanced equation for this reaction: 4Al(s) + 3O2(g) ==> 2Al2O3(s)
From this, note that 4 moles Al react with 3 moles O2 to produce 2 moles Al2O3.
Moles of Al used = 20.50 g Al x 1 mole Al/26.98 g = 0.7598 moles Al
Since there is excess O2, Al will be limiting and will determine moles of Al2O3 formed.
Moles Al2O3 formed = 0.7598 moles Al x 2 moles Al2O3/4 moles Al = 0.3799 moles Al2O3 formed theoretically.
Grams Al2O3 theoretically formed = 0.3799 moles x 101.96 g/mole = 38.74 g
% yield = 35.48 g/38.74 g (x100%) = 91.58%