The solution to your problem is as follows:
2 KClO3 → 2 KCl + 3 O2
MW of KCL = 39.1 + 35.35 = 74.6 g/mole
62.6/74.6 = 0.839 moles KCl produced
0.839 moles KCl x 3O2/2KCl x 32g/mole O2 = 40.27g of O2 was produced
Therefore, there are 40.27g of O2 produced during the reaction.
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