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 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between sulfur dioxide and oxygen to form sulfur trioxide is:
\[ 2 \, SO_2(g) + O_2(g) \rightleftharpoons 2 \, SO_3(g) \]

### Step 2: Determine initial moles
Initially, we have:
- Moles of \( SO_2 = 5 \)
- Moles of \( O_2 = 5 \)
- Moles of \( SO_3 = 0 \)

### Step 3: Calculate the amount of \( SO_2 \) used
We are told that 60% of \( SO_2 \) is used up at equilibrium. Therefore, the moles of \( SO_2 \) that react can be calculated as:
\[ \text{Moles of } SO_2 \text{ used} = 60\% \text{ of } 5 = \frac{60}{100} \times 5 = 3 \text{ moles} \]

### Step 4: Calculate remaining moles of \( SO_2 \)
The remaining moles of \( SO_2 \) after the reaction is:
\[ \text{Remaining } SO_2 = 5 - 3 = 2 \text{ moles} \]

### Step 5: Determine moles of \( O_2 \) used
From the balanced equation, we see that 2 moles of \( SO_2 \) react with 1 mole of \( O_2 \). Therefore, if 3 moles of \( SO_2 \) are used, the moles of \( O_2 \) used can be calculated as:
\[ \text{Moles of } O_2 \text{ used} = \frac{3}{2} = 1.5 \text{ moles} \]

### Step 6: Calculate remaining moles of \( O_2 \)
The remaining moles of \( O_2 \) after the reaction is:
\[ \text{Remaining } O_2 = 5 - 1.5 = 3.5 \text{ moles} \]

### Step 7: Calculate moles of \( SO_3 \) formed
From the balanced equation, 2 moles of \( SO_2 \) produce 2 moles of \( SO_3 \). Therefore, if 3 moles of \( SO_2 \) are used, the moles of \( SO_3 \) formed will be:
\[ \text{Moles of } SO_3 = 3 \text{ moles} \]

### Step 8: Calculate total moles at equilibrium
Now, we can calculate the total number of moles at equilibrium:
\[ \text{Total moles} = \text{Remaining } SO_2 + \text{Remaining } O_2 + \text{Moles of } SO_3 \]
\[ \text{Total moles} = 2 + 3.5 + 3 = 8.5 \text{ moles} \]

### Final Answer
The total number of moles of \( SO_2, O_2, \) and \( SO_3 \) in the vessel at equilibrium is **8.5 moles**.

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