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 chemical equation for the formation of ammonia from nitrogen and hydrogen is:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]

### Step 2: Identify the Stoichiometric Ratios
From the balanced equation, we can see that:
- 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

### Step 3: Assume Equal Volumes
Let’s assume we have equal volumes of nitrogen and hydrogen. Let the volume of each gas be \( X \) liters.

### Step 4: Calculate Moles of Each Gas
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the number of moles can be calculated as:
- Moles of N₂ = \( \frac{X}{22.4} \)
- Moles of H₂ = \( \frac{X}{22.4} \)

### Step 5: Apply Stoichiometric Ratios
According to the stoichiometry of the reaction:
- For every 1 mole of N₂, 3 moles of H₂ are required.
- Therefore, the moles of H₂ required for \( \frac{X}{22.4} \) moles of N₂ is:
\[ 3 \times \frac{X}{22.4} = \frac{3X}{22.4} \]

### Step 6: Compare Available Moles with Required Moles
Now we need to compare the available moles of H₂ with the required moles:
- Available moles of H₂ = \( \frac{X}{22.4} \)
- Required moles of H₂ = \( \frac{3X}{22.4} \)

### Step 7: Determine the Limiting Reagent
Since the available moles of H₂ (\( \frac{X}{22.4} \)) are less than the required moles of H₂ (\( \frac{3X}{22.4} \)), hydrogen (H₂) is the limiting reagent.

### Conclusion
Thus, the limiting reagent in this reaction is hydrogen (H₂).