Simple chemistry question using gas laws

How many moles of KClO3 at STP will be needed to produce 22.4 L of O2 gas according to the following equation:
2KClO3(s) => 2KCl(s) + 3O2(g)

The answer is .67 but I am not sure how to get it. I tried using PV =nRT

1. Use PV = nRT to find out how many moles of O2 gas were produced.

n = PV/RT
= plug in givens
= 0.9997 mol

2. Now do some simple mol balancing ....
0.9997 mol O2 x 2 mol KClO3/3mol O2
= 0.66649 mol

Therefore, there is 0.67 moles of KClO3.