Calculate the volume of oxygen produced in the decomposition of 5 moles of KClO3 at STP

Answer 1

2KClO3 → 2KCl + 3O2
so 5 moles of KClO3 will produce 7.5 moles of O2
since each mole occupies 22.4L, that means you get
7.5 * 22.4 = 168 L of O2