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