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, follow these steps:

### Step 1: Write the balanced chemical equation.
The conversion of oxygen to ozone can be represented by the following balanced equation:
\[ 3 O_2 \rightarrow 2 O_3 \]

### Step 2: Calculate the number of moles of oxygen (O2).
Using the ideal gas law, we can find the number of moles of O2 using the formula:
\[ \text{Number of moles} = \frac{\text{Volume}}{22.4 \, \text{L/mol}} \]
Given that the volume of oxygen is 67.2 L:
\[ \text{Number of moles of } O_2 = \frac{67.2 \, \text{L}}{22.4 \, \text{L/mol}} = 3 \, \text{moles} \]

### 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, if we have 3 moles of O2:
\[ \text{Moles of } O_3 = \frac{2}{3} \times \text{Moles of } O_2 = \frac{2}{3} \times 3 = 2 \, \text{moles} \]

### Step 4: Calculate the actual yield of ozone (O3) based on the conversion extent.
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).
To find the mass of O3, we use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
The molar mass of O3 (ozone) is:
\[ 3 \times 16 \, \text{g/mol} = 48 \, \text{g/mol} \]
Thus, the mass of O3 produced is:
\[ \text{Mass of } O_3 = 0.3 \, \text{moles} \times 48 \, \text{g/mol} = 14.4 \, \text{grams} \]

### Final Answer:
The mass of ozone that can be prepared is **14.4 grams**.

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