The volume of O₂ obtained by the complete decomposition of 2.45g of KClO₃ (Formula Wt.: 122.5) at STP is:$2{\mathrm{KClO}}_3\to 2\mathrm{KCl}+3{\mathrm{O}}_2$

The chemical equation for decomposition of KClO3 can be written as:

$2{\mathrm{KClO}}_3\to 2\mathrm{KCl}+3{\mathrm{O}}_2$

245 g of KClO₃ gives 67.2 L of O₂ at STP.

for 2.45 g of KClO₃ gives 0.672 L of O₂ at STP=672 mL of O₂ at STP