5 moles of `SO_(2)` and 5 moles of `O_(2)` react in a closed vessel. At equilibrium 60% of the `SO_(2)` is consumed . The total number of gaseous moles`(SO_(2),O_(2)andSO_(3))` in the vessel is :-

Options

A. 5.1
B. 3.9
C. 10.5
D. 8.5

Answer

To solve the problem, we need to analyze the reaction and the changes in the number of moles of the reactants and products at equilibrium.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
The reaction between sulfur dioxide (SO₂) and oxygen (O₂) to form sulfur trioxide (SO₃) can be represented as:
\[ 2 \, SO_2 + O_2 \rightarrow 2 \, SO_3 \]

2. **Identify initial moles:**
We start with 5 moles of SO₂ and 5 moles of O₂. At the beginning (t=0):
- Moles of SO₂ = 5
- Moles of O₂ = 5
- Moles of SO₃ = 0

3. **Determine the amount of SO₂ consumed:**
According to the problem, 60% of SO₂ is consumed at equilibrium. Therefore, the amount of SO₂ consumed is:
\[ \text{Moles of SO₂ consumed} = 60\% \text{ of } 5 = 0.6 \times 5 = 3 \text{ moles} \]

4. **Calculate remaining moles of SO₂:**
The remaining moles of SO₂ after the reaction is:
\[ \text{Remaining moles of SO₂} = 5 - 3 = 2 \text{ moles} \]

5. **Determine moles of O₂ consumed:**
From the balanced equation, for every 2 moles of SO₂ consumed, 1 mole of O₂ is consumed. Therefore, the moles of O₂ consumed when 3 moles of SO₂ are consumed is:
\[ \text{Moles of O₂ consumed} = \frac{3}{2} = 1.5 \text{ moles} \]

6. **Calculate remaining moles of O₂:**
The remaining moles of O₂ after the reaction is:
\[ \text{Remaining moles of O₂} = 5 - 1.5 = 3.5 \text{ moles} \]

7. **Calculate moles of SO₃ produced:**
According to the balanced equation, 2 moles of SO₃ are produced for every 2 moles of SO₂ consumed. Therefore, the moles of SO₃ produced from 3 moles of SO₂ consumed is:
\[ \text{Moles of SO₃ produced} = 3 \text{ moles} \]

8. **Calculate total moles at equilibrium:**
Now, we can find the total number of moles in the vessel at equilibrium:
\[ \text{Total moles} = \text{Remaining moles of SO₂} + \text{Remaining moles of O₂} + \text{Moles of SO₃ produced} \]
\[ \text{Total moles} = 2 + 3.5 + 3 = 8.5 \text{ moles} \]

### Final Answer:
The total number of gaseous moles (SO₂, O₂, and SO₃) in the vessel at equilibrium is **8.5 moles**.

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