How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 19.0 grams of O2?

Reaction: 2Al + 3O2 → 2Al2O3

4 Al + 3 O2 → 2 Al2O3

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al

(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =

10.1 g O2 left over