The mole ratio of oxygen (O2) to KClO3 in the decomposition reaction is 3 : 2.
Moles of oxygen produced
= (Moles of KClO3 used) × 3/2
= 5 mol × 3/2
= 7.5 mol
Volume of 1 mol of oxygen at STP = 22.4 L
Volume of 7.5 mol of oxygen = 22.4 L × 7.5
= 168 L (answer)