Potassium chlorate decomposes to form potassium chloride and oxygen. Balancefd equation is as follows; 2KClO₃ --> 2KCl + 3O₂ stoichiometry of KClO₃ to O₂ is 2:3 At STP, 1 mol of any gas occupies a volume of 22.4 L. Number of moles in 22.4 L is - 1 mol Therefore in 11.2 L - 1/22.4 x 11.2 = 0.5 mol When 3 mol of O₂ are formed - 2 mol of KClO₃ react therefore when 0.5 mol of O₂ are formed - 2/3 x 0.5 = 0.33 mol of KClO₃ reacted Therefore 0.33 mol of KClO₃ are required the mass of KClO₃ required - 0.33 mol x 122.5 g/mol = 40.43 g
