Using the molar volume in calculations with chemical equations
2 marks · NormalCalculate the volume, in dm³, of chlorine gas produced at rtp from the complete decomposition of 2.7 g of potassium chlorate, KClO₃. The balanced equation is:
(Ar: K = 39, Cl = 35.5, O = 16; molar volume = 24 dm³)
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0.336 dm^3 (±0.01)
Common mistakes: Not using correct molar mass (Mr) for KClO3; Incorrect gas ratio
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Marking points
M1 · 1 mark · Correctly finds moles of KClO3 (2.7 / 122.5)
A1 · 1 mark · Correct calculation for O2 volume (using ratio, conversion; accept 0.336 dm³ to 2/3 sig fig) (follow-through)
Worked solution
Step 1: Calculate Mr of KClO3: 39 + 35.5 + (3×16) = 39 + 35.5 + 48 = 122.5
From the equation, 2 mol KClO3 produces 3 mol O2, so 0.02204 mol KClO3 produces (3/2)×0.02204 = 0.03306 mol O2.