Question

Solve problem:

Calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at STP. Molar mass of KClO₃ is 122.5 g mol^-1.

Answer 1

The molecular formula of potassium chlorate is KClO₃. Required chemical equation:

\[\ce{\underset{\text{[2 moles]}}{2KClO3}->2KCl + \underset{\text{[3 moles]}}{3O2} ^}\]

2 moles of KClO₃ = 2 × 122.5 = 245 g
3 moles of O₂ at STP occupy = (3 × 22.4 dm³) = 67.2 dm³
Thus, 245 g of potassium chlorate will liberate 67.2 dm³ of oxygen gas.
Let ‘x’ gram of KClO₃ liberate 6.72 dm³ of oxygen gas at S.T.P.
∴ x = `(245xx6.72)/67.2` = 24.5 g

Mass of potassium chlorate required = 24.5 g