Question

How many moles of potassium chlorate are heated to produce \(22.4 \text{lit}\) of oxygen at STP ?

• A. \(\frac{1}{2} \text{mole}\)
• B. \(\frac{1}{3} \text{mole}\)
• C. \(\frac{1}{4} \text{mole}\)
• D. \(\frac{2}{3} \text{mole}\)

Hint:

Write the decomposition equation and note that 22.4 L of gas at STP is exactly 1 mole.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Potassium chlorate decomposes on heating to give potassium chloride and oxygen gas. We need the mole ratio from the balanced equation and the molar volume of a gas at STP.

Step 2: Key Formula or Approach:
1. Balanced equation: \(2\text{KClO}_3 \to 2\text{KCl} + 3\text{O}_2\).
2. At STP, 1 mole of any gas occupies \(22.4\) L.

Step 3: Detailed Explanation:
Moles of oxygen formed \(= \frac{22.4}{22.4} = 1\) mole.
From the equation, 3 moles of \(\text{O}_2\) come from 2 moles of \(\text{KClO}_3\).
So 1 mole of \(\text{O}_2\) needs \(\frac{2}{3}\) mole of \(\text{KClO}_3\).
\[ n(\text{KClO}_3) = 1 \times \frac{2}{3} = \frac{2}{3}\ \text{mol} \]
Options A, B and C come from using a wrong mole ratio (for example 1:2 or 1:3 instead of 2:3), so they do not match the balanced equation.

Final Answer:
Heating \(\frac{2}{3}\) mole of potassium chlorate gives \(22.4\) L of oxygen at STP, which is option (D). \[ \boxed{\frac{2}{3}\ \text{mole}} \]