Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s). The equation for the reaction is
2KClO3 ---> 2KCl+3O2
Calculate how many grams of O2(g) can be produced from heating 72.0 grams of KClO3(s).

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2 KClO3 → 2 KCl + 3 O2

(19.7 g KClO3 / (122.5495 g KClO3/mol) x (3 mol O2 / 2 mol KClO3) x (31.99886 g O2/mol) = 7.72 g O2
I hope this is correct. Boi, this is so confusing