Hi

Since 2 moles of KClO3 yields 2 moles of KCl and 3 moles of O2

Therefore, the mole ratio for the reaction is 2:2:1

According to the information provided in the question,

By using ideal gas law, 22.4l = 1 mole

Then, 6.72l= 1×6.72/22.4 = 0.3 moles

As per the mole ratio, 0.3 moles of oxygen to be liberated, we need 0.3×2=0.6 moles of KClO3

Moles= mass/molar mass

Molar mass of potassium chlorate is 122.5g/mol that is given and moles is 0.6

Hence, mass of KClO3 required = 122.5g/mol×0.6 moles = 73.5g

Thank you