In the reaction, `2SO_(2) +O_(2) to 2SO_(3)` when 1 mole of `SO_(2)` and 1 mole of `O_(2)` are made to react to completion

Options

A. All the oxygen will be consumed
B. 1.0 mole of `SO_(3)` will be produced
C. 0.5 mole of `SO_(2)` is remained
D. All of these

Answer

To solve the problem, we need to analyze the reaction given:

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

### Step 1: Identify the initial amounts of reactants
We start with 1 mole of \(\text{SO}_2\) and 1 mole of \(\text{O}_2\).

### Step 2: Determine the stoichiometry of the reaction
From the balanced equation, we see that:
- 2 moles of \(\text{SO}_2\) react with 1 mole of \(\text{O}_2\) to produce 2 moles of \(\text{SO}_3\).

### Step 3: Calculate the limiting reactant
Since we have 1 mole of \(\text{SO}_2\):
- To react with 1 mole of \(\text{SO}_2\), we need \( \frac{1}{2} \) mole of \(\text{O}_2\) (because 2 moles of \(\text{SO}_2\) require 1 mole of \(\text{O}_2\)).
- We have 1 mole of \(\text{O}_2\), which is more than enough to react with 1 mole of \(\text{SO}_2\).

Thus, \(\text{SO}_2\) is the limiting reactant.

### Step 4: Calculate the amount of products formed
Since 1 mole of \(\text{SO}_2\) reacts, according to the stoichiometry of the reaction:
- 1 mole of \(\text{SO}_2\) produces 1 mole of \(\text{SO}_3\).

### Step 5: Determine the remaining amounts of reactants
After the reaction:
- All of the 1 mole of \(\text{SO}_2\) is consumed.
- \(\frac{1}{2}\) mole of \(\text{O}_2\) is consumed (since 1 mole of \(\text{SO}_2\) reacts with \( \frac{1}{2} \) mole of \(\text{O}_2\)).
- Therefore, the remaining amount of \(\text{O}_2\) is:
\[ 1 - 0.5 = 0.5 \text{ moles of } \text{O}_2 \]

### Summary of Results
- Moles of \(\text{SO}_3\) produced: 1 mole
- Moles of \(\text{SO}_2\) remaining: 0 moles
- Moles of \(\text{O}_2\) remaining: 0.5 moles

### Conclusion
The correct answer is:
- **Option B:** 1 mole of \(\text{SO}_3\) will be produced.