When 40 c c of slightly moist hydrogen chloride gas is mixed with 20 c c of ammonia gas the final volume of gas left at the same temperature and pressure will be
`NH_(3)(g)HCl(g)toNH_(4)Cl(s)`

Options

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

Answer

To solve the problem, we need to analyze the reaction between hydrogen chloride (HCl) and ammonia (NH3) and determine the final volume of gas left after the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between hydrogen chloride gas and ammonia gas can be represented as:
\[ NH_3(g) + HCl(g) \rightarrow NH_4Cl(s) \]
This indicates that ammonia gas reacts with hydrogen chloride gas to form solid ammonium chloride.

2. **Determine Initial Volumes**:
- Volume of HCl = 40 cc
- Volume of NH3 = 20 cc

3. **Identify the Limiting Reactant**:
In this reaction, both gases react in a 1:1 molar ratio. Therefore, we need to determine which reactant will be consumed first.
- Since we have 20 cc of NH3 and 40 cc of HCl, NH3 is the limiting reactant because it will be completely consumed first.

4. **Calculate the Volume of Gases Consumed**:
- The 20 cc of NH3 will react with 20 cc of HCl to form 20 cc of NH4Cl (which is solid and does not contribute to the gas volume).
- Therefore, the volume of HCl that reacts is also 20 cc.

5. **Calculate the Remaining Volume of Gases**:
- Remaining volume of HCl after the reaction:
\[ \text{Remaining HCl} = \text{Initial HCl} - \text{Consumed HCl} = 40 \text{ cc} - 20 \text{ cc} = 20 \text{ cc} \]
- Remaining volume of NH3 after the reaction:
\[ \text{Remaining NH3} = \text{Initial NH3} - \text{Consumed NH3} = 20 \text{ cc} - 20 \text{ cc} = 0 \text{ cc} \]

6. **Final Volume of Gas**:
The final volume of gas left at the same temperature and pressure is the remaining volume of HCl:
\[ \text{Final Volume} = \text{Remaining HCl} + \text{Remaining NH3} = 20 \text{ cc} + 0 \text{ cc} = 20 \text{ cc} \]

### Conclusion:
The final volume of gas left after the reaction is **20 cc**.