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 gas (N₂) and hydrogen gas (H₂) react to form ammonia (NH₃), we can follow these steps:

### Step 1: Write the balanced chemical equation.
The reaction between nitrogen and hydrogen to form ammonia is represented by the balanced equation:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

### Step 2: Identify the mole ratio from the balanced equation.
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: Analyze the given condition.
The problem states that equal volumes of N₂ and H₂ are provided. Since gases at the same temperature and pressure occupy equal volumes, we can assume that the number of moles of N₂ and H₂ is also equal when measured under the same conditions.

### Step 4: Determine the limiting reagent.
Since 1 mole of N₂ requires 3 moles of H₂ for the reaction, if we have 1 mole of N₂, we would need 3 moles of H₂. However, since we only have equal volumes of N₂ and H₂, we have only 1 mole of H₂ available to react with 1 mole of N₂.

### Step 5: Conclusion on limiting reagent.
Given that we need 3 moles of H₂ to completely react with 1 mole of N₂, and we only have 1 mole of H₂, hydrogen (H₂) is the limiting reagent in this reaction.

### Final Answer:
The limiting reagent is **H₂**.

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