How many moles of potassium chlorate need to be heated to produce 5.2 litres of oxygen at NTP?

Options

A. 0.12
B. 1.2
C. 2.1
D. 0.21

Explanation

To solve the problem of how many moles of potassium chlorate (KClO3) need to be heated to produce 5.2 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 5.2 liters using the formula:

\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at NTP}} = \frac{5.2 \, \text{liters}}{22.4 \, \text{liters/mole}} \]

Calculating this gives:

\[ \text{Number of moles of } O_2 = \frac{5.2}{22.4} \approx 0.2321 \, \text{moles} \]

### 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:

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

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

### Step 3: Relate the moles of KClO3 to the moles of O2 produced

From the balanced equation, we can determine the ratio of moles of KClO3 to moles of O2:

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

### Step 4: Calculate the number of moles of KClO3 needed for 0.2321 moles of O2

Using the ratio from the balanced equation, we can set up the following proportion to find the moles of KClO3 required:

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

Substituting the value of moles of O2:

\[ \text{Moles of } KClO_3 = 0.2321 \times \frac{2}{3} \approx 0.1547 \, \text{moles} \]

### Final Answer

Approximately 0.155 moles of potassium chlorate (KClO3) need to be heated to produce 5.2 liters of oxygen at NTP.

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