How many moles of potassium chlorate need to be heated to produce 6.2 litres of oxygen at NTP?

Options

A. 0.18
B. 1.8
C. 0.36
D. 0.6

Explanation

To solve the problem of how many moles of potassium chlorate (KClO3) need to be heated to produce 6.2 liters of oxygen (O2) at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Determine the volume of oxygen produced in moles
At NTP, 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the number of moles of oxygen in 6.2 liters using the formula:

\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at NTP}} = \frac{6.2 \text{ L}}{22.4 \text{ L/mol}} \]

Calculating this gives:

\[ \text{Number of moles of } O_2 = \frac{6.2}{22.4} \approx 0.276 \text{ moles} \]

### Step 2: Write the balanced chemical equation
The decomposition of potassium chlorate (KClO3) can be represented by the following balanced equation:

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

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

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

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

To find the moles of KClO3 required to produce 0.276 moles of O2, we can set up a proportion:

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

Substituting the value we found for moles of O2:

\[ \text{Moles of KClO}_3 = \frac{2}{3} \times 0.276 \approx 0.184 \text{ moles} \]

### Step 4: Final answer
Thus, the number of moles of potassium chlorate that need to be heated to produce 6.2 liters of oxygen at NTP is approximately **0.18 moles**.

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