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Question

You are given a 2.00 g mixture of the compounds potassium chlorate and sodium chloride. Heating this mixture will cause the potassium chlorate to decompose as follows. 2KClO3 = 2KCl + 3O2. The NaCl will only melt and recrystallize. After heating the muxture twice to obtain a constant weight, the final weight of the mixture was 1.75 g. What was the percent by weight of potaesium chlorate in the original mixture of potassium chlorate and sodium chloride

Answer 1

Look at the balanced equation: 2KClO₃ ==> 2KCl + 3O₂

NOTE that oxygen gas is being formed, and that will be lost to the atmosphere. So, whatever loss in mass we find, it due to the oxygen formed.

molar mass KClO₃ = 123 g / mol

molar mass O₂ = 32 g / mol

mass lost on heating = 2.00 g - 1.75 g = 0.25 g. This is the mass of O₂

moles of O₂ = 0.25 g x 1 mol / 32 g = 0.00781 mols O₂

moles KClO₃ present = 0.00781 mols O₂ x 2 mol KClO₃ / 3 mols O₂ = 0.00521 mols KClO₃

mass KClO₃ = 0.00521 mols x 123 g / mol = 0.641 g KClO₃

% by weight KClO₃ = 0.641 g / 2.00 g (x100%) = 32.0%