Taking one mole of `N_(2(g))` and excess of `H_(2)(g)`, under suitable conditions, formation of `NH_(3)` is completed when `NH_(3)(g)` has mole fraction of `0.5`, The initial moles of `H_(2)` are?

Explanation

To solve the problem of determining the initial moles of \( H_2 \) when 1 mole of \( N_2 \) is used to produce \( NH_3 \) with a mole fraction of 0.5, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the formation of ammonia from nitrogen and hydrogen is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

### Step 2: Set up initial conditions
Let’s denote:
- Initial moles of \( N_2 \) = 1 mole
- Initial moles of \( H_2 \) = \( A \) moles (unknown)
- Initial moles of \( NH_3 \) = 0 moles

### Step 3: Determine moles at equilibrium
At the completion of the reaction, the moles of each component will be:
- Moles of \( N_2 \) = \( 1 - 1 \) (since 1 mole of \( N_2 \) reacts completely) = 0 moles
- Moles of \( H_2 \) = \( A - 3 \) (3 moles of \( H_2 \) are consumed for every mole of \( N_2 \) reacted)
- Moles of \( NH_3 \) = \( 2 \) (produced from 1 mole of \( N_2 \))

### Step 4: Calculate total moles at equilibrium
The total moles at equilibrium can be expressed as:
\[ \text{Total moles} = \text{Moles of } N_2 + \text{Moles of } H_2 + \text{Moles of } NH_3 = 0 + (A - 3) + 2 = A - 1 \]

### Step 5: Use the mole fraction to find \( A \)
The mole fraction of \( NH_3 \) is given as 0.5. The mole fraction is defined as:
\[ \text{Mole fraction of } NH_3 = \frac{\text{Moles of } NH_3}{\text{Total moles}} = \frac{2}{A - 1} \]
Setting this equal to 0.5 gives us the equation:
\[ \frac{2}{A - 1} = 0.5 \]

### Step 6: Solve for \( A \)
Cross-multiplying to solve for \( A \):
\[ 2 = 0.5(A - 1) \]
\[ 2 = 0.5A - 0.5 \]
\[ 2 + 0.5 = 0.5A \]
\[ 2.5 = 0.5A \]
\[ A = \frac{2.5}{0.5} = 5 \]

### Conclusion
The initial moles of \( H_2 \) are \( 5 \) moles.