Equal volume of `N_(2)` and `H_(2)` react to form ammonia under suitable condition then the limiting reagent is

Options

A. `H_(2)`
B. `N_(2)`
C. `NH_(3)`
D. No one reactant is limiting reagent

Answer

To determine the limiting reagent when equal volumes of nitrogen (N₂) and hydrogen (H₂) react to form ammonia (NH₃), 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:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Understand the stoichiometry
From the balanced equation, we can see that:
- 1 mole of nitrogen (N₂) reacts with 3 moles of hydrogen (H₂) to produce 2 moles of ammonia (NH₃).

### Step 3: Determine the relationship between moles of reactants
Since we are given equal volumes of N₂ and H₂, we can apply Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules. Therefore, if we take:
- Volume of N₂ = Volume of H₂ = V

This implies:
- Moles of N₂ = Moles of H₂

Let’s denote the moles of N₂ as \( x \) and the moles of H₂ as \( x \).

### Step 4: Calculate the required moles of H₂ for the given moles of N₂
From the stoichiometry of the reaction:
- For every 1 mole of N₂, 3 moles of H₂ are required.

Thus, for \( x \) moles of N₂, the moles of H₂ required will be:
\[ 3 \times x \]

### Step 5: Compare available moles of H₂ with required moles
Since we have only \( x \) moles of H₂ available, we can compare:
- Required moles of H₂ = \( 3x \)
- Available moles of H₂ = \( x \)

### Step 6: Identify the limiting reagent
Since the available moles of H₂ (which is \( x \)) are less than the required moles of H₂ (which is \( 3x \)), hydrogen (H₂) will be the limiting reagent.

### Conclusion
The limiting reagent in this reaction is:
\[ \text{H}_2 \]