How many moles of potassium chlorate need to be heated to produce 14.2 litres of oxygen at NTP?

Options

A. 1.6
B. 0.36
C. 3.6
D. 2.6

Explanation

To solve the problem of how many moles of potassium chlorate (KClO3) need to be heated to produce 14.2 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, one mole of any gas occupies 22.4 liters. We can use this information to calculate the number of moles of oxygen in 14.2 liters.

\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at NTP}} = \frac{14.2 \, \text{liters}}{22.4 \, \text{liters/mole}} \]

Calculating this gives:

\[ \text{Number of moles of } O_2 = \frac{14.2}{22.4} \approx 0.634 \, \text{moles} \]

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

\[ 2 \, KClO_3 \rightarrow 2 \, KCl + 3 \, O_2 \]

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

### Step 3: Calculate the number of moles of KClO3 required
From the balanced equation, we can establish the ratio of moles of KClO3 to moles of O2 produced:

\[ \frac{2 \, \text{moles of } KClO_3}{3 \, \text{moles of } O_2} \]

Now, we can use this ratio to find the number of moles of KClO3 needed to produce 0.634 moles of O2:

\[ \text{Moles of } KClO_3 = \frac{2}{3} \times \text{Moles of } O_2 \]

Substituting the value of moles of O2:

\[ \text{Moles of } KClO_3 = \frac{2}{3} \times 0.634 \approx 0.423 \, \text{moles} \]

### Final Answer
Thus, approximately 0.423 moles of potassium chlorate (KClO3) need to be heated to produce 14.2 liters of oxygen at NTP.

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