How many moles of potassium chlorate need to be heated to produce 8.2 litres of oxygen at NTP?

Options

A. 0.15
B. 0.24
C. 2.4
D. None of these

Explanation

To solve the problem of how many moles of potassium chlorate (KClO3) need to be heated to produce 8.2 liters of oxygen (O2) at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Understand the volume of oxygen produced per mole
At NTP, 1 mole of any gas occupies 22.4 liters. Therefore, we can use this information to find out how many moles of oxygen are in 8.2 liters.

### Step 2: Calculate the moles of oxygen
Using the formula:
\[ \text{Moles of O}_2 = \frac{\text{Volume of O}_2}{\text{Volume occupied by 1 mole at NTP}} = \frac{8.2 \, \text{liters}}{22.4 \, \text{liters/mole}} \]
Calculating this gives:
\[ \text{Moles of O}_2 = \frac{8.2}{22.4} \approx 0.365 \, \text{moles} \]
This can be approximated to 0.36 moles.

### Step 3: Write the balanced chemical equation
The decomposition of potassium chlorate is represented by the following balanced equation:
\[ 2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, \text{O}_2 \]
From this equation, we see that 2 moles of KClO3 produce 3 moles of O2.

### Step 4: Determine the mole ratio
From the balanced equation, we can establish the ratio of KClO3 to O2:
\[ \frac{2 \, \text{moles of KClO}_3}{3 \, \text{moles of O}_2} \]

### Step 5: Calculate the moles of KClO3 needed
Now, we can use the mole ratio to find out how many moles of KClO3 are needed to produce 0.36 moles of O2:
\[ \text{Moles of KClO}_3 = \frac{2}{3} \times \text{Moles of O}_2 \]
Substituting the value of moles of O2:
\[ \text{Moles of KClO}_3 = \frac{2}{3} \times 0.36 \approx 0.24 \, \text{moles} \]

### Conclusion
Thus, the number of moles of potassium chlorate required to produce 8.2 liters of oxygen at NTP is approximately **0.24 moles**.

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