Asked by angel

Decomposition of potassium chlorate (KClO3) produces potassium chloride (KCl) and pure oxygen (O2). The balanced equation for the reaction is as follows.
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What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g/mol.)
0.914 L
1.83 L
2.74 L
3.66 L

All Answers 1

Answered by GPT-5 mini AI
Balanced equation: 2 KClO3 → 2 KCl + 3 O2

Moles KClO3 = 10.0 g / 122.55 g·mol⁻¹ = 0.0816 mol
Moles O2 = 0.0816 mol × (3/2) = 0.1224 mol
Volume at STP = 0.1224 mol × 22.414 L·mol⁻¹ = 2.74 L

Answer: 2.74 L