Question:
determine the amount of potassium chlorate required to produce1.12 L of oxygen gas at STP
Answer:
2KClO3 → 2KCl + 3O2
2mol KClO3 will produce 3 mol O2
At STP 1 mol O2 = 22.4L
6.72L = 6.72/22.4 = 0.3 mol O2
This will require 0.2 mol KClO3
Mass KClO3 required = 0.2 x 122.5 = 24.5g
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