How many moles of potassium chlorate need to be heated to produce 16.2 litres of oxygen at NTP?

Options

A. 0.21
B. 0.42
C. 0.31
D. 4.2

Explanation

To solve the problem of how many moles of potassium chlorate (KClO3) need to be heated to produce 16.2 liters of oxygen (O2) at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Calculate the number of moles of oxygen produced.
At NTP, 1 mole of any gas occupies 22.4 liters. Therefore, we can use this relationship to find the number of moles of oxygen in 16.2 liters.

\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Volume occupied by 1 mole at NTP}} = \frac{16.2 \text{ L}}{22.4 \text{ L/mol}} \]

Calculating this gives:

\[ \text{Number of moles of } O_2 = \frac{16.2}{22.4} \approx 0.724 \text{ moles} \]

### Step 2: Write the balanced chemical equation for the decomposition of potassium chlorate.
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 the equation, we can see that 2 moles of KClO3 produce 3 moles of O2.

### Step 3: Determine the moles of KClO3 required to produce the calculated moles of O2.
Using the stoichiometry from the balanced equation, we can set up a ratio:

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

Now, we can find out how many moles of KClO3 are needed to produce 0.724 moles of O2:

\[ \text{Moles of KClO}_3 = \left(\frac{2 \text{ moles of KClO}_3}{3 \text{ moles of } O_2}\right) \times 0.724 \text{ moles of } O_2 \]

Calculating this gives:

\[ \text{Moles of KClO}_3 = \frac{2}{3} \times 0.724 \approx 0.4827 \text{ moles} \]

### Final Answer:
Approximately **0.48 moles** of potassium chlorate (KClO3) are needed to produce 16.2 liters of oxygen at NTP.

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