Calculate the amount of KClO3 which on thermal decomposition gives ‘X’ vol. of O2, which is the volume required for combustion of 24 g. of carbon.
2KClO3 ⟶ 2KCl + 3O2 ;
C + O2 ⟶ CO2.
[K = 39, Cl = 35.5, O = 16, C = 12].

C      +     O₂    {\overset{ \triangle }{\longrightarrow}}     CO₂

12 g of C gives 22.4 lit of CO₂

∴ 24 g of C will give 22.4\over12 × 24

= 44.8 lit. of CO₂

Hence, 24 g of C will give 44.8 of CO₂

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

67.2 lit of O₂ is produced by 245 g of KClO₃

∴ 44.8 lit of O₂ is produced by 245\over67.2 × 44.8

= 163.33 g

Hence, the amount of KClO₃ is 163.33 g.