`KClO_(3)` on heating decomposes to give KCl and `O_(2)`. What is the volume of `O_(2)` at N.T.P liberated by 0.1 mole of `KClO_(3)` ?

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the balanced chemical equation for the decomposition of KClO₃.
The decomposition of potassium chlorate (KClO₃) can be represented by the following balanced equation:
\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]

### Step 2: Determine the mole ratio between KClO₃ and O₂.
From the balanced equation, we can see that:
- 2 moles of KClO₃ produce 3 moles of O₂.

### Step 3: Calculate the amount of O₂ produced from 0.1 mole of KClO₃.
Using the mole ratio from the balanced equation, we can find the moles of O₂ produced from 0.1 mole of KClO₃:
\[ \text{Moles of O}_2 = \left(0.1 \text{ moles KClO}_3\right) \times \left(\frac{3 \text{ moles O}_2}{2 \text{ moles KClO}_3}\right) = 0.15 \text{ moles O}_2 \]

### Step 4: Calculate the volume of O₂ at NTP.
At Normal Temperature and Pressure (NTP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of O₂ produced can be calculated as follows:
\[ \text{Volume of O}_2 = \text{Moles of O}_2 \times 22.4 \text{ L/mole} = 0.15 \text{ moles} \times 22.4 \text{ L/mole} = 3.36 \text{ L} \]

### Final Answer:
The volume of O₂ liberated by 0.1 mole of KClO₃ at NTP is **3.36 liters**.

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