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 step by step, we need to analyze the reaction and the volumes of the gases involved.

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2SO_2 (g) + O_2 (g) \rightarrow 2SO_3 (g) \]

### Step 2: Identify initial volumes
From the problem, we have:
- Volume of \( SO_2 \) = 5.0 L
- Volume of \( O_2 \) = 3.0 L

### Step 3: Determine the stoichiometric requirements
According to the balanced equation:
- 2 volumes of \( SO_2 \) react with 1 volume of \( O_2 \).
- Therefore, 5.0 L of \( SO_2 \) will require:
\[ \text{Required } O_2 = \frac{5.0 \, \text{L} \, SO_2}{2} = 2.5 \, \text{L} \, O_2 \]

### Step 4: Identify the limiting reagent
We have 3.0 L of \( O_2 \) available, which is more than the 2.5 L required to react with 5.0 L of \( SO_2 \). Thus, \( SO_2 \) is the limiting reagent.

### Step 5: Calculate the final volumes after the reaction
- Since \( SO_2 \) is the limiting reagent, it will be completely consumed:
- Final volume of \( SO_2 \) = 0 L
- The volume of \( O_2 \) that reacts is 2.5 L, so the remaining volume of \( O_2 \) is:
\[ \text{Remaining } O_2 = 3.0 \, \text{L} - 2.5 \, \text{L} = 0.5 \, \text{L} \]
- The volume of \( SO_3 \) produced is equal to the volume of \( SO_2 \) consumed, which is 5.0 L (since 2 volumes of \( SO_2 \) produce 2 volumes of \( SO_3 \)).

### Step 6: Calculate the total final volume
The total final volume of the reaction mixture is the sum of the volumes of the remaining \( O_2 \) and the produced \( SO_3 \):
\[ \text{Total final volume} = \text{Volume of } SO_3 + \text{Remaining } O_2 = 5.0 \, \text{L} + 0.5 \, \text{L} = 5.5 \, \text{L} \]

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

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