How many moles of potassium chlorate need to be heated to produce 1.2 litres of oxygen at NTP?

Options

A. 0.01
B. 0.1
C. 0.03
D. 0.3

Explanation

To solve the problem of how many moles of potassium chlorate (KClO3) need to be heated to produce 1.2 liters of oxygen (O2) at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Understand the volume of oxygen produced
At NTP, 1 mole of any gas occupies 22.4 liters. Therefore, we can use this information to calculate how many moles of oxygen are present in 1.2 liters.

### Calculation:
\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at NTP}} = \frac{1.2 \, \text{liters}}{22.4 \, \text{liters/mole}} = 0.05357 \, \text{moles} \approx 0.05 \, \text{moles} \]

### Step 2: Write the decomposition reaction of potassium chlorate
The decomposition of potassium chlorate can be represented by the following balanced chemical equation:
\[ 2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, O_2 \]
From this equation, we can see that 2 moles of KClO3 produce 3 moles of O2.

### Step 3: Determine the mole ratio
From the balanced equation, we can establish the ratio of moles of KClO3 to moles of O2:
\[ \frac{2 \, \text{moles of KClO}_3}{3 \, \text{moles of } O_2} \]

### Step 4: Calculate the moles of KClO3 needed
Now, we can use the ratio to find out how many moles of KClO3 are needed to produce 0.05 moles of O2:
\[ \text{Moles of KClO}_3 = \left(\frac{2 \, \text{moles of KClO}_3}{3 \, \text{moles of } O_2}\right) \times 0.05 \, \text{moles of } O_2 \]
\[ \text{Moles of KClO}_3 = \frac{2}{3} \times 0.05 = 0.03333 \, \text{moles} \approx 0.03 \, \text{moles} \]

### Conclusion
Therefore, the number of moles of potassium chlorate that need to be heated to produce 1.2 liters of oxygen at NTP is approximately **0.03 moles**.

---