How many moles of potassium chlorate are needed to form 22.4 L oxygen at STP

Answer 1

Decomposition of potassium chlorate:

2KClO3 ==heat==> 2KCl + 3O2 ... balanced equation

At STP, 22.4 L O2 x 1 mole O2/22.4 L = 1 mole O2 formed

moles KClO3 needed = 1 mole O2 x 2 moles KClO3/3 moles O2 = 0.667 moles KClO3 needed