How many grams of potassium chlorate are needed to create 11.2 liters of oxygen gas at stp?

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 

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