Question:
Calculate the mass of potassium chlorate required to liberate \(6.72\) dm\({}^{3}\) of oxygen at STP.
[Molar mass of potassium chlorate \(=122.5\) g mol\({}^{-1}\)]
Calculate the mass of potassium chlorate required to liberate \(6.72\) dm\({}^{3}\) of oxygen at STP.
[Molar mass of potassium chlorate \(=122.5\) g mol\({}^{-1}\)]
[Molar mass of potassium chlorate \(=122.5\) g mol\({}^{-1}\)]
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Convert the volume of oxygen to moles at STP and use the 2:3 mole ratio from the decomposition equation.
Updated On: Nov 12, 2026
- \(12.25\) g
- \(24.5\) g
- \(36.75\) g
- \(49\) g
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Potassium chlorate decomposes on heating: \(2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2\).
Step 2: Moles of oxygen
At STP, 1 mol of gas occupies 22.4 dm\(^{3}\), so \(n_{\text{O}_2} = \dfrac{6.72}{22.4} = 0.3\) mol.
Step 3: Moles and mass of KClO3
From the equation, 3 mol O\(_2\) comes from 2 mol KClO\(_3\), so \(n = 0.3\times\dfrac{2}{3} = 0.2\) mol.
Mass \(= 0.2\times 122.5 = 24.5\) g.
Final Answer:
24.5 g of potassium chlorate is required. \[ \boxed{\text{(B) }24.5\text{ g}} \]
Potassium chlorate decomposes on heating: \(2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2\).
Step 2: Moles of oxygen
At STP, 1 mol of gas occupies 22.4 dm\(^{3}\), so \(n_{\text{O}_2} = \dfrac{6.72}{22.4} = 0.3\) mol.
Step 3: Moles and mass of KClO3
From the equation, 3 mol O\(_2\) comes from 2 mol KClO\(_3\), so \(n = 0.3\times\dfrac{2}{3} = 0.2\) mol.
Mass \(= 0.2\times 122.5 = 24.5\) g.
Final Answer:
24.5 g of potassium chlorate is required. \[ \boxed{\text{(B) }24.5\text{ g}} \]
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