Question

Calculate the mass, in grams, of O2 that would be produced following the reaction of 72.36 grams of potassium chlorate (KClO3).

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

Answer 1

how many moles KClO3 in 72.36g?
O2 produced will be 3/2 that many moles.
convert back to grams

Answer 2

2 KClO3(s) → 2 KCl(s) + 3 O2(g)
mols KClO3 = grams/molar mass = 72.36/122.55 = ? Check those numbers.
mols O2 produced = mols KClO3 x (3 mols O2/2 mols KClO3) = ?
Then grams O2 = mols O2 x molar mass O2 = ?
Post your work if you get stuck.