In the reaction 2KClO3(s) --> 2KCl(s) + 3O2(g) If 22.4 L of oxygen at STP are needed, how many moles of KClO3 must be decomposed?

Answer 1

1 mole of a gas at STP occupies a volume of 22.4 L. From the given reaction, 1 mole of O₂ is formed from the decomposition of KClO₃.

From the balanced chemical reaction, you can see that 3 moles of O₂ are formed from 2 moles of KClO₃. Using stoichiometry,

mol of KClO₃ = 1 mol O₂ x (2 mol KClO₃ / 3 mol O₂) = 0.7 mol KClO₃ (rounded to 1 significant figure)