MHT CET Medical202623 April 2026Morning ShiftChemistrySome Basic Concepts of ChemistryActual
Calculate the mass of potassium chlorate required to liberate 6.72 dm ^3 of oxygen at STP. [Molar mass of potassium chlorate = 122.5 g mol ⁻¹ ]
Options
- A12.25 g
- B24.5 g
- C36.75 g
- D49 g
Correct answer
B. 24.5 g
Step-by-step solution
The balanced chemical equation for the decomposition of potassium chlorate is: 2KClO₃ 2KCl + 3O₂ From the stoichiometry of the reaction, 2 moles of KClO₃ produce 3 moles of O₂ . At STP, the volume of 1 mole of an ideal gas is 22.4 dm ^3 . Volume of 3 moles of O₂ = 3 22.4 = 67.2 dm ^3 . Mass of 2 moles of KClO₃ = 2 122.5 = 245 g. Thus, 67.2 dm ^3 of O₂ is liberated by 245 g of KClO₃ . Mass of KClO₃ required to liberate 6.72 dm ^3 of O₂ = 245 67.2 6.72 = 24.5 g. Answer: 24.5 g