`N_(2)`(g) reacts with `H_(2)`(g) in either of the following ways depending upon supply of `H_(2)`(g) :
`N_(2)(g)+H_(2)(g) to N_(2)H_(2)(l)`
`N_(2)(g)+2H_(2)(g) to N_(2)H_(4)`(g)
If 5 L `N_(2)`(g) and 3 L `H_(2)`(g) are taken initially (at same temperature and pressure ), calculate the contraction in valume after the reaction (in L ).

Answer

To solve the problem, we need to analyze the two possible reactions of nitrogen gas (N₂) with hydrogen gas (H₂) and calculate the contraction in volume after the reaction.

### Step-by-Step Solution:

1. **Identify the Initial Volumes:**
- We have 5 L of N₂ and 3 L of H₂.
- Total initial volume (V_initial) = Volume of N₂ + Volume of H₂ = 5 L + 3 L = 8 L.

2. **Consider the First Reaction:**
- The first reaction is:
\[ N₂(g) + H₂(g) \rightarrow N₂H₂(l) \]
- According to this reaction:
- 1 mole of N₂ reacts with 1 mole of H₂ to produce 1 mole of liquid N₂H₂.
- Since the volume of a liquid is negligible compared to gases, the volume of N₂H₂ can be considered as 0 L.

3. **Calculate the Volume Change for the First Reaction:**
- For the first reaction, if all 3 L of H₂ react with N₂:
- 3 L of H₂ will consume 3 L of N₂ (since 1:1 ratio).
- Remaining N₂ = 5 L - 3 L = 2 L.
- Final volume (V_final) after the first reaction:
\[ V_{final} = \text{Remaining N₂} + \text{Volume of liquid N₂H₂} = 2 L + 0 L = 2 L. \]
- Contraction in volume:
\[ \text{Contraction} = V_{initial} - V_{final} = 8 L - 2 L = 6 L. \]

4. **Consider the Second Reaction:**
- The second reaction is:
\[ N₂(g) + 2H₂(g) \rightarrow N₂H₄(g) \]
- According to this reaction:
- 1 mole of N₂ reacts with 2 moles of H₂ to produce 1 mole of N₂H₄.
- Since we only have 3 L of H₂, we can determine the limiting reagent:
- 3 L of H₂ can react with 1.5 L of N₂ (since 2 L of H₂ is needed for 1 L of N₂).
- Therefore, we can use 1.5 L of N₂ and 3 L of H₂.

5. **Calculate the Volume Change for the Second Reaction:**
- After the reaction:
- 1.5 L of N₂ reacts with 3 L of H₂ to produce 1.5 L of N₂H₄.
- Remaining N₂ = 5 L - 1.5 L = 3.5 L.
- Final volume (V_final) after the second reaction:
\[ V_{final} = \text{Remaining N₂} + \text{Volume of N₂H₄} = 3.5 L + 1.5 L = 5 L. \]
- Contraction in volume:
\[ \text{Contraction} = V_{initial} - V_{final} = 8 L - 5 L = 3 L. \]

### Conclusion:
- The contraction in volume after the first reaction is 6 L.
- The contraction in volume after the second reaction is 3 L.