The conversion of oxygen to ozone occurs to the extent of 15% only. The mass of ozone that can be prepared from 67.2 L of oxygen at 1 atm and 273 K will be :

Options

A. 14.4 g
B. 96 g
C. 640 g
D. 64 g

Answer

To solve the problem of determining the mass of ozone (O₃) that can be prepared from 67.2 L of oxygen (O₂) at 1 atm and 273 K, we can follow these steps:

### Step 1: Calculate the number of moles of oxygen (O₂)
Using the ideal gas law, we know that at standard temperature and pressure (STP: 0°C and 1 atm), 1 mole of gas occupies 22.4 L.

\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{67.2 \text{ L}}{22.4 \text{ L/mol}} = 3 \text{ moles} \]

### Step 2: Write the balanced chemical equation for the conversion of O₂ to O₃
The balanced equation for the conversion of oxygen to ozone is:

\[ 3 O_2 \rightarrow 2 O_3 \]

### Step 3: Determine the theoretical yield of ozone (O₃)
From the balanced equation, we see that 3 moles of O₂ produce 2 moles of O₃. Therefore, the number of moles of O₃ produced from 3 moles of O₂ is:

\[ \text{Moles of } O_3 = \frac{2}{3} \times \text{Moles of } O_2 = \frac{2}{3} \times 3 = 2 \text{ moles} \]

### Step 4: Calculate the actual yield of ozone (O₃) based on the conversion efficiency
Since the conversion of O₂ to O₃ occurs to the extent of 15%, we calculate the actual moles of O₃ produced:

\[ \text{Actual moles of } O_3 = 2 \text{ moles} \times \frac{15}{100} = 0.3 \text{ moles} \]

### Step 5: Calculate the mass of ozone (O₃)
The molar mass of ozone (O₃) is approximately 48 g/mol. Therefore, the mass of O₃ produced is:

\[ \text{Mass of } O_3 = \text{Moles of } O_3 \times \text{Molar mass of } O_3 = 0.3 \text{ moles} \times 48 \text{ g/mol} = 14.4 \text{ g} \]

### Final Answer
The mass of ozone that can be prepared from 67.2 L of oxygen at 1 atm and 273 K is **14.4 g**.

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