At constant T and P,5.0 L of `SO_2` are reacted with 3.0 L of `O_2` according to the following equation `2SO_2 (g)+O_2(g)rarr2SO_3(g)` . The volume of the reaction mixture at the completion of the reaction is

Options

A. 0.5 L
B. 8.0 L
C. 5.5 L
D. 5 L

Answer

To solve the problem, we need to analyze the reaction and determine the volumes of the reactants and products involved.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g) \]

2. **Identify the initial volumes of the reactants:**
- Volume of SO₂ = 5.0 L
- Volume of O₂ = 3.0 L

3. **Determine the stoichiometric ratios:**
From the balanced equation, we see that:
- 2 volumes of SO₂ react with 1 volume of O₂.
- Therefore, 1 volume of SO₂ requires 0.5 volumes of O₂.

4. **Calculate the amount of O₂ required for the available SO₂:**
- For 5.0 L of SO₂, the required volume of O₂ is:
\[ \text{Required O}_2 = \frac{5.0 \, \text{L SO}_2}{2} = 2.5 \, \text{L O}_2 \]

5. **Identify the limiting reagent:**
- We have 3.0 L of O₂ available, but only 2.5 L is needed to react with 5.0 L of SO₂.
- Therefore, SO₂ is the limiting reagent because it will be completely consumed first.

6. **Calculate the volumes of the products formed:**
- According to the balanced equation, 2 volumes of SO₂ produce 2 volumes of SO₃.
- Thus, 5.0 L of SO₂ will produce:
\[ \text{Volume of SO}_3 = 5.0 \, \text{L} \]

7. **Calculate the remaining volume of O₂ after the reaction:**
- Initial volume of O₂ = 3.0 L
- Volume of O₂ consumed = 2.5 L
- Remaining volume of O₂:
\[ \text{Remaining O}_2 = 3.0 \, \text{L} - 2.5 \, \text{L} = 0.5 \, \text{L} \]

8. **Calculate the total volume of the reaction mixture at completion:**
- Total volume at completion = Volume of SO₃ produced + Remaining O₂
\[ \text{Total Volume} = 5.0 \, \text{L} + 0.5 \, \text{L} = 5.5 \, \text{L} \]

### Final Answer:
The volume of the reaction mixture at the completion of the reaction is **5.5 L**.

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