At STP 10 L of `H_(2)S` was reacted with 10 L of `SO_(2)`. The volume of gas remaining after the reaction is complete would be

Options

A. 5 L
B. 10 L
C. 15 L
D. 20 L

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the balanced chemical equation.
The reaction between hydrogen sulfide (H₂S) and sulfur dioxide (SO₂) can be represented as follows:

\[ 2 \text{H}_2\text{S} + \text{SO}_2 \rightarrow 2 \text{S} + 2 \text{H}_2\text{O} \]

### Step 2: Determine the mole ratio from the balanced equation.
From the balanced equation, we can see that:
- 2 moles of H₂S react with 1 mole of SO₂.

### Step 3: Convert the volumes of gases to moles.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. However, since we are dealing with volumes directly (and at STP, volumes can be treated as moles), we can use the volumes directly:
- Volume of H₂S = 10 L
- Volume of SO₂ = 10 L

### Step 4: Determine the limiting reagent.
According to the stoichiometry:
- 10 L of H₂S would require \( \frac{10 \text{ L H}_2\text{S}}{2} = 5 \text{ L SO}_2 \) to completely react.
- 10 L of SO₂ would require \( 10 \text{ L SO}_2 \times 2 = 20 \text{ L H}_2\text{S} \) to completely react.

Since we only have 10 L of H₂S, it is the limiting reagent. This means that all of the H₂S will be consumed in the reaction.

### Step 5: Calculate the remaining volume of SO₂ after the reaction.
Since 5 L of SO₂ is consumed in the reaction with 10 L of H₂S, we can find the remaining volume of SO₂:
- Initial volume of SO₂ = 10 L
- Volume of SO₂ consumed = 5 L
- Remaining volume of SO₂ = \( 10 \text{ L} - 5 \text{ L} = 5 \text{ L} \)

### Step 6: Conclusion
The volume of gas remaining after the reaction is complete is 5 L of SO₂.

### Final Answer:
The volume of gas remaining after the reaction is complete would be **5 L**.

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