Equal volumes of `H_(2) and Cl_(2)` are mixed. How will the volume of the mixtuer change after the reaction?

Options

A. Unchanged
B. Reduced to half
C. Increases two fold
D. None of these

Answer

To solve the problem of how the volume of the mixture changes after the reaction between equal volumes of \( H_2 \) and \( Cl_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between hydrogen gas (\( H_2 \)) and chlorine gas (\( Cl_2 \)) can be represented as:
\[ H_2(g) + Cl_2(g) \rightarrow 2 HCl(g) \]

2. **Determine Initial Volumes**:
Let's assume we mix equal volumes of \( H_2 \) and \( Cl_2 \). For this example, we can take:
- Volume of \( H_2 = 100 \, \text{ml} \)
- Volume of \( Cl_2 = 100 \, \text{ml} \)

Therefore, the total initial volume \( V_1 \) is:
\[ V_1 = 100 \, \text{ml} + 100 \, \text{ml} = 200 \, \text{ml} \]

3. **Calculate the Volume of Products**:
According to the stoichiometry of the reaction, 1 mole of \( H_2 \) reacts with 1 mole of \( Cl_2 \) to produce 2 moles of \( HCl \). This means:
- 100 ml of \( H_2 \) will produce 200 ml of \( HCl \).

4. **Determine Final Volume**:
After the reaction, the volume of \( HCl \) produced is:
\[ V_{\text{final}} = 200 \, \text{ml} \]

5. **Compare Initial and Final Volumes**:
Now, we compare the initial volume \( V_1 \) and the final volume \( V_{\text{final}} \):
- Initial Volume \( V_1 = 200 \, \text{ml} \)
- Final Volume \( V_{\text{final}} = 200 \, \text{ml} \)

6. **Conclusion**:
Since the initial volume and the final volume are the same, we conclude that the volume of the mixture remains unchanged after the reaction.

### Final Answer:
The volume of the mixture remains unchanged.

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