4 moles each of `SO_(2) "and" O_(2)` gases are allowed to react to form `SO_(3)` in a closed vessel. At equilibrium 25% of `O_(2)` is used up. The total number of moles of all the gases at equilibrium is

Options

A. 6.5
B. 7
C. 8
D. 2

Explanation

To solve the problem step by step, we will analyze the reaction and the changes in the number of moles of the gases involved.

### 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 written as:
\[ 2 \, \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \, \text{SO}_3(g) \]

2. **Initial Moles:**
Initially, we have:
- Moles of SO₂ = 4
- Moles of O₂ = 4
- Moles of SO₃ = 0

3. **Determine the Change in Moles:**
We are told that 25% of O₂ is used up at equilibrium. Since we started with 4 moles of O₂, the amount used up is:
\[ \text{Amount of O}_2 \text{ used} = 25\% \text{ of } 4 = 1 \text{ mole} \]

4. **Calculate Moles at Equilibrium:**
If 1 mole of O₂ is used, we can find the changes in the moles of each substance:
- Moles of O₂ at equilibrium = Initial moles - Used moles = \(4 - 1 = 3\)
- According to the stoichiometry of the reaction, for every 1 mole of O₂ used, 2 moles of SO₂ are consumed. Therefore, the moles of SO₂ used = 2 × 1 = 2 moles.
- Moles of SO₂ at equilibrium = Initial moles - Used moles = \(4 - 2 = 2\)
- Moles of SO₃ formed = 2 moles (since 2 moles of SO₂ produce 2 moles of SO₃).

5. **Total Moles at Equilibrium:**
Now we can calculate the total number of moles of all gases at equilibrium:
\[ \text{Total moles} = \text{Moles of SO}_2 + \text{Moles of O}_2 + \text{Moles of SO}_3 \]
\[ \text{Total moles} = 2 + 3 + 2 = 7 \]

### Final Answer:
The total number of moles of all the gases at equilibrium is **7 moles**.