How many moles of potassium chlorate need to be heated to produce 10.2 litres of oxygen at NTP?

Explanation

To determine how many moles of potassium chlorate (KClO3) need to be heated to produce 10.2 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 gas occupies 22.4 liters. Therefore, we can find the number of moles of oxygen in 10.2 liters using the formula:

\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at NTP}} = \frac{10.2 \, \text{liters}}{22.4 \, \text{liters/mole}} \]

Calculating this gives:

\[ \text{Number of moles of } O_2 = \frac{10.2}{22.4} \approx 0.4554 \, \text{moles} \]

### 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:

\[ 2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, O_2 \]

### Step 3: Determine the mole ratio between KClO3 and O2.
From the balanced equation, we see that 2 moles of KClO3 produce 3 moles of O2. This gives us the ratio:

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

### Step 4: Calculate the number of moles of KClO3 needed to produce the calculated moles of O2.
Using the mole ratio, we can set up the following proportion to find the moles of KClO3 required to produce 0.4554 moles of O2:

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

Substituting the value of moles of O2:

\[ \text{Moles of KClO}_3 = \left( \frac{2}{3} \right) \times 0.4554 \approx 0.3036 \, \text{moles} \]

### Final Answer:
Approximately 0.30 moles of potassium chlorate (KClO3) need to be heated to produce 10.2 liters of oxygen at NTP.

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