How many moles of potassium chlorate need to be heated to produce 11.2 litres of oxygen at NTP?

Options

A. `1/2`
B. `2/3`
C. `1/3`
D. `1/4`

Explanation

To solve the question of how many moles of potassium chlorate (KClO3) need to be heated to produce 11.2 liters of oxygen (O2) at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Determine the number of moles of oxygen produced
At NTP, 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the number of moles of oxygen in 11.2 liters using the formula:

\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at NTP}} = \frac{11.2 \, \text{liters}}{22.4 \, \text{liters/mole}} = 0.5 \, \text{moles} \]

### Step 2: Write the balanced chemical equation for the decomposition of potassium chlorate
The balanced equation for the decomposition of potassium chlorate is:

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

### Step 3: Relate moles of KClO3 to moles of O2
From the balanced equation, we see that 2 moles of KClO3 produce 3 moles of O2. Therefore, we can set up a ratio to find out how many moles of KClO3 are needed to produce 0.5 moles of O2:

\[ \frac{2 \, \text{moles KClO}_3}{3 \, \text{moles O}_2} = \frac{x \, \text{moles KClO}_3}{0.5 \, \text{moles O}_2} \]

### Step 4: Solve for x (moles of KClO3)
Cross-multiplying gives:

\[ 2 \times 0.5 = 3 \times x \]

This simplifies to:

\[ 1 = 3x \implies x = \frac{1}{3} \, \text{moles of KClO}_3 \]

### Conclusion
Thus, to produce 11.2 liters of oxygen at NTP, we need \( \frac{1}{3} \) moles of potassium chlorate (KClO3).

### Final Answer
\(\frac{1}{3}\) moles of KClO3 are required.

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