If 73.5 grams of fes2 is allowed to react with 44 grams of o2 according to the following equation, how many grams of fe2o3 are produced? fes2 + o2 → fe2o3 + so2

Balanced equation:

4 FeS₂ + 11 O₂ --> 2 Fe₂O₃ + 8 SO₂

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Find the limiting reactant (the one that produces the least amount of product) :

(73.5 g FeS₂) * (1 mole FeS₂ / 119.975 g FeS₂) * (2 moles Fe₂O₃ / 4 moles FeS2) * (159.6882 g Fe₂O₃ / 1 mole Fe₂O₃)

= 48.9 g Fe₂O₃ …………. to 3 sig figs

(44 g O₂) * (1 mole O₂ / 31.9988 g O₂) * (2 moles Fe₂O₃ / 11 moles O₂) *  

(159.6882 g Fe₂O₃ / 1 mole Fe₂O₃)

= 40 g Fe₂O₃ ………………. to 2 sig figs

O₂ is the limiting reactant and the theoretical yield is 44 g Fe₂O₃

the answer is d.) potassium