How many moles of potassium chlorate should be decomposed completely to obtain 67.2 litres of oxygen at STP

The reaction is 2KClO3→2KCl+3O2. The decomposition of two moles of potassium chlorate gives three moles of oxygen at STP. At STP, one mole of oxygen occupies 22.4 L, hence, 3 moles of oxygen will occupy 67.2 L. Thus, to obtain 67.2 L of oxygen at STP, 2 moles of potassium chlorate should be decomposed.