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
