Question

4. Calculate the volume of oxygen produced in the decomposition of 5 moles of KCIO3 at STP?​

Answer 1

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)

Answer 2

2KClO3=2KCl + 3O2

n(O2)=7.5 moles

V(O2)=7.5*22.4=168 litres