First, you must write the correctly balanced equation for the reaction taking place:
4Al(s) + 3O2(g) ==> 2Al2O3(s) ... balanced equation
Next, find the limiting reactant. One easy way is to divide the moles of each reactant by the corresponding coefficient in the balanced equation.
Al: 13.0 g Al x 1 mol Al / 26.98 g = 0.482 mols Al (÷4 -> 0.12)
O2: 21.0 g O2 x 1 mol O2 / 32 g = 0.656 mols O2 (÷3 -> 0.219)
Al is LIMITING since 0.12 is less than 0.219
Finally, compute the mass of Al2O3 using the limiting reactant, stoichiometry and dimensional analysis:
0.482 mols Al x 2 mol Al2O3 / 4 mol Al x 102 g Al2O3 / mol = 24.6 g Al2O3 produced