KClO3  {\overset{ MnO2 }{\longrightarrow}}  2KCl + 3O2
(i) Calculate the mass of KClO3 required to produce 6.72 lit of O2 at s.t.p.
[K = 39, Cl = 35.5, O = 16]
(ii) Calculate the number of moles of O2 in the above volume and also the number of molecules.

KClO₃  {\overset{ MnO2 }{\longrightarrow}}  2KCl + 3O₂

245 g             149 g   67.2 lit

67.2 lit. of O₂ is produced by 245 g of KClO₃

∴ 6.72 lit of O₂ will be obtained from 245\over 67.2 × 6.72 = 24.5 g.

(ii) 22.4 lit = 1 mole

∴ 6.72 lit = 1\over22.4 × 6.72 = 0.3 moles

1 mole = 6.023 × 10²³ molecules

∴ 0.3 moles = 0.3 × 6.023 × 10²³ molecules.