1 mol of `N_2` and 4 mol of `H_2` are allowed to react in a vessel and after reaction, `H_2O` is added. Aqueous solution required 1 mol of HCI for neutralization. Mol fraction of `H_2` in the mixture after reaction is :

Options

A. `1/2`
B. `5/6`
C. `1/3`
D. `3/4`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Write the balanced chemical reaction
The reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to form ammonia (NH₃) can be represented as:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Determine the initial moles of reactants
From the question, we have:
- Moles of \( N_2 \) = 1 mol
- Moles of \( H_2 \) = 4 mol

### Step 3: Identify the limiting reactant
According to the stoichiometry of the reaction, 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \). Therefore, for 1 mole of \( N_2 \), we need:
\[ 3 \text{ moles of } H_2 \]
Since we have 4 moles of \( H_2 \), \( N_2 \) is the limiting reactant.

### Step 4: Calculate the moles of \( H_2 \) that react
Since \( N_2 \) is the limiting reactant, it will completely react. The amount of \( H_2 \) that reacts with 1 mole of \( N_2 \) is:
\[ 3 \text{ moles of } H_2 \]
Thus, the moles of \( H_2 \) that will remain after the reaction is:
\[ \text{Remaining } H_2 = 4 \text{ moles} - 3 \text{ moles} = 1 \text{ mole} \]

### Step 5: Calculate the moles of \( NH_3 \) produced
From the balanced equation, 1 mole of \( N_2 \) produces 2 moles of \( NH_3 \). Therefore, the moles of \( NH_3 \) produced will be:
\[ 2 \text{ moles of } NH_3 \]

### Step 6: Determine the total moles after the reaction
After the reaction, we have:
- Remaining \( H_2 \) = 1 mole
- \( N_2 \) = 0 moles (since it is completely consumed)
- \( NH_3 \) = 2 moles

Total moles in the mixture after the reaction:
\[ \text{Total moles} = 1 \text{ (H}_2\text{)} + 0 \text{ (N}_2\text{)} + 2 \text{ (NH}_3\text{)} = 3 \text{ moles} \]

### Step 7: Calculate the mole fraction of \( H_2 \)
The mole fraction of \( H_2 \) is given by the formula:
\[ \text{Mole fraction of } H_2 = \frac{\text{Moles of } H_2}{\text{Total moles}} \]
Substituting the values:
\[ \text{Mole fraction of } H_2 = \frac{1 \text{ mole}}{3 \text{ moles}} = \frac{1}{3} \]

### Step 8: Final Answer
The mole fraction of \( H_2 \) in the mixture after the reaction is:
\[ \frac{1}{3} \]

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