How many moles of potassium chlorate need to be heated to produce 22.4 litres of oxygen at NTP?

Options

A. 0.1
B. 0.66
C. 1
D. 6

Explanation

To determine how many moles of potassium chlorate (KClO3) need to be heated to produce 22.4 liters of oxygen (O2) at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Understand the relationship between moles and volume at NTP
At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, if we want to produce 22.4 liters of oxygen, we can deduce that:

\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Volume occupied by 1 mole at NTP}} = \frac{22.4 \, \text{liters}}{22.4 \, \text{liters/mole}} = 1 \, \text{mole} \]

### Step 2: Write the balanced chemical equation
The decomposition of potassium chlorate can be represented by the following balanced equation:

\[ 2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, \text{O}_2 \]

From this equation, we can see that 2 moles of KClO3 produce 3 moles of O2.

### Step 3: Determine the moles of KClO3 required for 1 mole of O2
From the balanced equation, we can establish the ratio of KClO3 to O2:

\[ \frac{2 \, \text{moles of KClO}_3}{3 \, \text{moles of } O_2} \]

This means that for every 3 moles of O2 produced, 2 moles of KClO3 are required. To find out how many moles of KClO3 are needed to produce 1 mole of O2, we can set up a proportion:

\[ \text{Moles of KClO}_3 = \frac{2}{3} \times \text{Moles of } O_2 \]

Substituting 1 mole for O2:

\[ \text{Moles of KClO}_3 = \frac{2}{3} \times 1 = \frac{2}{3} \, \text{moles} \]

### Step 4: Conclusion
Thus, to produce 22.4 liters of oxygen at NTP, we need:

\[ \frac{2}{3} \, \text{moles of KClO}_3 \approx 0.67 \, \text{moles} \]

### Final Answer
You need approximately **0.67 moles of potassium chlorate (KClO3)** to produce 22.4 liters of oxygen at NTP.

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