If 'x' `dm^3` of` N_2` react with 10`dm^3` of` H_2` to form `NH_3` under suitable conditions, then

Options

A. `H_2` is limiting reagent
B. `N_2` is limiting reagent
C. Both reactants are limiting reagents
D. `NH_3` is limiting reagent

Answer

To solve the problem step by step, we will analyze the reaction between nitrogen (N₂) and hydrogen (H₂) to form ammonia (NH₃) and determine the limiting reagent based on the volumes given.

### Step 1: Write the balanced chemical equation
The reaction between nitrogen and hydrogen to form ammonia can be represented by the balanced equation:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]
This indicates that 1 volume of nitrogen reacts with 3 volumes of hydrogen to produce 2 volumes of ammonia.

### Step 2: Identify the volumes given
From the question, we have:
- Volume of nitrogen (N₂) = \( x \, \text{dm}^3 \)
- Volume of hydrogen (H₂) = \( 10 \, \text{dm}^3 \)

### Step 3: Determine the stoichiometric ratio
According to the balanced equation:
- 1 volume of N₂ reacts with 3 volumes of H₂.
- Therefore, for \( x \, \text{dm}^3 \) of N₂, the required volume of H₂ would be \( 3x \, \text{dm}^3 \).

### Step 4: Set up the equation based on the given volume of H₂
Since we have 10 dm³ of H₂ available, we can set up the equation:
\[ 3x = 10 \]
To find \( x \), we solve for \( x \):
\[ x = \frac{10}{3} \approx 3.33 \, \text{dm}^3 \]

### Step 5: Identify the limiting reagent
Now, we compare the available volumes:
- We have \( x \approx 3.33 \, \text{dm}^3 \) of N₂.
- We need \( 3 \times 3.33 = 10 \, \text{dm}^3 \) of H₂ for complete reaction.

Since we have exactly 10 dm³ of H₂, both reactants are present in the correct stoichiometric ratio, and neither is in excess. However, if we were to use the values directly, we would find that nitrogen is the limiting reagent because it is present in a smaller volume compared to the stoichiometric requirement.

### Conclusion
Thus, the limiting reagent in this reaction is nitrogen (N₂) with a volume of approximately 3.33 dm³.