5 moles of `SO_(2)` and 5 moles of `O_(2)` are allowed to react to form `SO_(3)` in a closed vessel. At the equilibrium stage `60%` of `SO_(2)` is used up. The total number of moles of `SO_(2), O_(2)` and `SO_(3)` in the vessel now is

Options

A. `10.0`
B. 8.5
C. 10.5
D. 3.9

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:
\[ 2 \, \text{SO}_2 + \text{O}_2 \rightarrow 2 \, \text{SO}_3 \]

2. **Initial Moles**:
At the start of the reaction, we have:
- Moles of SO₂ = 5
- Moles of O₂ = 5
- Moles of SO₃ = 0

3. **Determine the Change in Moles**:
According to the problem, 60% of SO₂ is used up.
- 60% of 5 moles of SO₂ = \(0.6 \times 5 = 3\) moles of SO₂ are consumed.

4. **Calculate Remaining Moles of SO₂**:
The remaining moles of SO₂ at equilibrium will be:
\[ \text{Remaining SO}_2 = 5 - 3 = 2 \, \text{moles} \]

5. **Calculate Moles of O₂ Used**:
From the balanced equation, for every 2 moles of SO₂ that react, 1 mole of O₂ is consumed. Therefore, for 3 moles of SO₂:
\[ \text{Moles of O}_2 \, \text{used} = \frac{3}{2} = 1.5 \, \text{moles} \]

6. **Calculate Remaining Moles of O₂**:
The remaining moles of O₂ at equilibrium will be:
\[ \text{Remaining O}_2 = 5 - 1.5 = 3.5 \, \text{moles} \]

7. **Calculate Moles of SO₃ Formed**:
According to the stoichiometry of the reaction, 3 moles of SO₂ produce 3 moles of SO₃. Therefore, the moles of SO₃ formed will be equal to the moles of SO₂ consumed:
\[ \text{Moles of SO}_3 = 3 \, \text{moles} \]

8. **Total Moles at Equilibrium**:
Now, we can calculate the total number of moles at equilibrium:
\[ \text{Total moles} = \text{Remaining SO}_2 + \text{Remaining O}_2 + \text{SO}_3 \]
\[ \text{Total moles} = 2 + 3.5 + 3 = 8.5 \, \text{moles} \]

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

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