The decomposition of potassium chlorate yields oxygen gas. If the yield is 85%, how many grams of KClO3 are needed to produce 15.0 L of O2?

2KClO3 (s) - - > 2KCl (s) + 3 O2 (g)

2KClO3 = 2KCl + 3O2.

10 L are 10/22.4 moles as 1 mole of gas at STP occupies 22.4 L so 0.446 moles but you only get 95% yield so you need to make more than 0.446 moles - - you need 0.446/0.95 = 0.470 moles.

Two moles KClO3 give 3 moles O2 so you need 0.470 * (2/3) moles KClO3 = 0.313 moles KClO3 and that has molar mass = 39 + 35.5 + 3*16 = 122.5 g so in g 0.313*122.5 = 38.4 g KClO3