Under similar conditions of pressure and temperature, 40 ml of slightly moist hydrogen chloride gas is mixed with 20 ml of ammonia gas, the final volume of gas at the same temperature and pressure will be

Options

A. 100 ml
B. 20 ml
C. 40 ml
D. 60 ml

Explanation

To solve the problem, we need to determine the final volume of gas after mixing 40 ml of hydrogen chloride (HCl) with 20 ml of ammonia (NH₃) under the same conditions of temperature and pressure.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between hydrogen chloride and ammonia can be represented as:
\[ \text{HCl (g)} + \text{NH}_3 (g) \rightarrow \text{NH}_4\text{Cl (s)} \]
Here, hydrogen chloride and ammonia react to form solid ammonium chloride.

2. **Determine the Volume of Reactants**:
- Volume of HCl = 40 ml
- Volume of NH₃ = 20 ml

3. **Identify the Limiting Reagent**:
According to the stoichiometry of the reaction, 1 mole of HCl reacts with 1 mole of NH₃. Since we have:
- 40 ml of HCl (which can be considered as 40 units)
- 20 ml of NH₃ (which can be considered as 20 units)

We can see that NH₃ is the limiting reagent because we have less of it (20 ml) compared to HCl (40 ml).

4. **Calculate the Volume of Products**:
Since NH₃ is the limiting reagent, it will determine the amount of product formed. For every 1 mole of NH₃, 1 mole of HCl is consumed, and 1 mole of NH₄Cl is produced. However, NH₄Cl is a solid and does not contribute to the volume of gas.

5. **Calculate the Final Volume of Gas**:
After the reaction, all of the NH₃ (20 ml) will react with an equal volume of HCl (20 ml), leaving:
- Remaining HCl = 40 ml - 20 ml = 20 ml
- Volume of gas after the reaction = Remaining HCl = 20 ml

### Final Answer:
The final volume of gas at the same temperature and pressure will be **20 ml**.