Answer:
This is a limiting reagent problem. the way to figure it out is to do the stoichiometry for each amount given, calculating the amount of Oxygen produced. The smallest amount is the answer because that reactant "ran out".
for KO2
2.49g KO2 x 1 mole KO2 x 3 mole O2 x 32.00 g O2 = 0.841 g O2
71.10 g KO2 4 mole KO2 1 mole O2
for CO2
4.44g CO2 x 1 mole CO2 x 3 mole O2 x 32.00g O2 = 4.84 g O2
44.01 g CO2 2 mole CO2 1 mole O2
Therefore the answer is 0.841g O2 since it is the smallest amount produced meaning KO2 is the limiting reagent.