How many moles of potassium chlorate should be decomposed completely to obtain 67.2 litres of oxygen at STP?

Options

A. 3
B. 4
C. 1
D. 2

Explanation

To find out how many moles of potassium chlorate (KClO3) should be decomposed to obtain 67.2 liters of oxygen (O2) at STP, we can follow these steps:

### Step 1: Write the balanced chemical equation for the decomposition of potassium chlorate.
The balanced equation for the decomposition of potassium chlorate is:
\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]

### Step 2: Determine the relationship between moles of KClO3 and O2.
From the balanced equation, we see that:
- 2 moles of KClO3 produce 3 moles of O2.

### Step 3: Calculate the number of moles of O2 produced from the given volume.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, to find the number of moles of O2 in 67.2 liters, we use the formula:
\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{67.2 \text{ L}}{22.4 \text{ L/mol}} = 3 \text{ moles} \]

### Step 4: Use stoichiometry to find the moles of KClO3 required.
From the balanced equation, we know that:
- 3 moles of O2 are produced from 2 moles of KClO3.

To find the moles of KClO3 needed to produce 3 moles of O2, we set up a proportion:
\[ \text{Moles of KClO}_3 = \frac{2 \text{ moles KClO}_3}{3 \text{ moles O}_2} \times 3 \text{ moles O}_2 = 2 \text{ moles KClO}_3 \]

### Final Answer:
Therefore, to obtain 67.2 liters of oxygen at STP, **2 moles of potassium chlorate (KClO3)** should be decomposed.

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