1 mol of `N_(2)` and 4 mol of `H_(2)` are allowed to react in a sealed container and after the reaction some water is introduced in it. The aqueous solution formed required 1 L of 1 M HCl for neutralization. Calculate the mole fraction of the gaseous product in the mixture after the reaction.

Explanation

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

### Step 1: Write the balanced chemical equation
The reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to form ammonia (NH₃) can be represented by the balanced equation:

\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Determine the limiting reactant
We start with 1 mole of N₂ and 4 moles of H₂. According to the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

- For 1 mole of N₂, we need 3 moles of H₂.
- We have 4 moles of H₂ available.

Since we have enough H₂ to react with the available N₂, N₂ is the limiting reactant.

### Step 3: Calculate the amount of NH₃ produced
From the balanced equation, 1 mole of N₂ produces 2 moles of NH₃. Therefore, from 1 mole of N₂, we can produce:

\[ 2 \text{ moles of } NH_3 \]

### Step 4: Calculate the remaining moles of H₂
Since 1 mole of N₂ reacts with 3 moles of H₂, the amount of H₂ that reacts with 1 mole of N₂ is 3 moles. We started with 4 moles of H₂, so the remaining moles of H₂ after the reaction will be:

\[ 4 \text{ moles of } H_2 - 3 \text{ moles of } H_2 = 1 \text{ mole of } H_2 \]

### Step 5: Calculate the remaining moles of N₂
Since all of the N₂ is consumed in the reaction, the remaining moles of N₂ will be:

\[ 1 \text{ mole of } N_2 - 1 \text{ mole of } N_2 = 0 \text{ moles of } N_2 \]

### Step 6: Calculate the total moles of gases after the reaction
After the reaction, we have:
- 0 moles of N₂
- 1 mole of H₂
- 2 moles of NH₃

Total moles of gases in the container:

\[ 0 + 1 + 2 = 3 \text{ moles} \]

### Step 7: Calculate the mole fraction of the gaseous product (NH₃)
The mole fraction of NH₃ is given by the formula:

\[ \text{Mole fraction of } NH_3 = \frac{\text{Moles of } NH_3}{\text{Total moles of gases}} = \frac{2}{3} \]

### Final Answer
The mole fraction of the gaseous product (NH₃) in the mixture after the reaction is:

\[ \text{Mole fraction of } NH_3 = \frac{2}{3} \approx 0.67 \]

---