Find the mass of potassium chloride required to liberate 5.6 ~dm ^3 of oxygen gas at STP ? (molar mass of KClO ₃=122.5 ~g / mol )

Step-by-step solution

The decomposition of potassium chlorate yields potassium chloride and oxygen gas, described by the balanced chemical equation: 2 KClO ₃(s) 2 KCl (s) + 3 O ₂(g) Given 5.6 dm ^3 of oxygen gas liberated at STP, where the molar volume is 22.4 dm ^3/ mol , the amount of oxygen is 5.6 22.4 = 0.25 mol . Using stoichiometry from the reaction, 3 mol of O ₂ require 2 mol of KClO ₃ . Thus, 0.25 mol of O ₂ requires 0.25 2 3 = 1 6 mol of KClO ₃ . The molar mass of KClO ₃ is 122.5 g/mol , so the required mass is 1 6 122.5 20.42