Find the mass of potassium chloride and volume of oxygen at STP liberated when 0.5 moles of potassium chlorate are heated.Also find the moles and mass of oxygen evolved.

To find the mass of potassium chloride and volume of oxygen at STP liberated when 0.5 moles of potassium chlorate are heated, we need to calculate the stoichiometry of the reaction between potassium chlorate and the products formed.

The balanced equation for the reaction is:
2KClO3(s) → 2KCl(s) + 3O2(g)

Step 1: Calculate the molar mass of potassium chlorate (KClO3):
- The molar mass of potassium (K) is approximately 39.10 g/mol.
- The molar mass of chlorine (Cl) is approximately 35.45 g/mol.
- The molar mass of oxygen (O) is approximately 16.00 g/mol.
- Since potassium chlorate has one potassium atom (K), one chlorine atom (Cl), and three oxygen atoms (O), the total molar mass is:
Molar mass of KClO3 = (39.10 g/mol) + (35.45 g/mol) + (16.00 g/mol x 3) = 122.55 g/mol

Step 2: Calculate the mass of potassium chlorate (KClO3):
- Given that we have 0.5 moles of KClO3.
- Use the formula: Mass of KClO3 = Moles of KClO3 x Molar mass of KClO3
Mass of KClO3 = 0.5 moles x 122.55 g/mol = 61.275 g

Step 3: Calculate the moles of oxygen (O2) evolved:
- According to the stoichiometry of the balanced equation, 2 moles of KClO3 produce 3 moles of O2.
- Use the formula: Moles of O2 = Moles of KClO3 x (3 moles of O2 / 2 moles of KClO3)
Moles of O2 = 0.5 moles x (3/2) = 0.75 moles

Step 4: Calculate the mass of oxygen (O2) evolved:
- Given that we have 0.75 moles of O2.
- The molar mass of oxygen (O2) is approximately 32.00 g/mol.
- Use the formula: Mass of O2 = Moles of O2 x Molar mass of O2
Mass of O2 = 0.75 moles x 32.00 g/mol = 24.00 g

Therefore, when 0.5 moles of potassium chlorate are heated, the mass of potassium chloride formed is 61.275 g, the volume of oxygen evolved at STP is not directly provided, but we can use the Ideal Gas Law to calculate it, and the moles and mass of oxygen evolved are 0.75 moles and 24.00 g, respectively.