Determine the volume of O2 (at STP) formed when 50.0 g of KClO3 decomposes according to the following reaction
Determine the volume of O2 (at STP) produced when 50.0 g of KClO3 decomposes according to the following reaction:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
Given that the molar mass of KClO3 is 122.55 g/mol, how can we calculate the volume of oxygen gas generated from this decomposition?
2 Answers
Feb 20, 2025
Take the given:
50.0 g KClO3
Multiply by 1 over mole weight:
50.0 g KClO3 x 1 mole / 122.55 g KClO3
Multiply by mole ratio (from equation):
50.0 g KClO3 x 1 mole / 122.55 g KClO3 x 3 mol O2/ 2 mol KClO3 = .612 mol O2
At STP, 1 mol of any gas occupies 22.4 L.
Set up a ratio:
.612 mol O2/ x Liters = 1 mol / 22.4 L
Cross mutliply
x= 13.7 L
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