In Haber process 30 litre of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only`50%` of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end ?

Options

A. 20 L ammonia , 10 L nitrogen , 30 L hydrogen
B. 20 L ammonia , 25 L nitrogen , 15 L hydrogen
C. 20 L ammonia , 20 L nitrogen , 20 L hydrogen
D. 10 L ammonia , 25 L nitrogen , 15 L hydrogen

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation for the Haber process.
The Haber process involves the reaction between nitrogen (N₂) and hydrogen (H₂) to form ammonia (NH₃). The balanced equation is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

### Step 2: Determine the initial amounts of reactants.
We are given:
- Volume of dihydrogen (H₂) = 30 L
- Volume of dinitrogen (N₂) = 30 L

### Step 3: Identify the limiting reagent.
From the balanced equation, we see that 1 mole of N₂ reacts with 3 moles of H₂. Therefore, for 30 L of N₂, we need:
\[ 30 \text{ L N₂} \times \frac{3 \text{ L H₂}}{1 \text{ L N₂}} = 90 \text{ L H₂} \]
Since we only have 30 L of H₂, H₂ is the limiting reagent.

### Step 4: Calculate the theoretical yield of NH₃.
According to the stoichiometry of the reaction:
- 3 moles of H₂ produce 2 moles of NH₃.
Thus, if we assume 100% yield, 30 L of H₂ would produce:
\[ \text{Volume of NH₃} = \frac{2}{3} \times 30 \text{ L} = 20 \text{ L} \]
However, we are told that the yield is only 50%, so the actual volume of NH₃ produced is:
\[ \text{Volume of NH₃} = 50\% \times 20 \text{ L} = 10 \text{ L} \]

### Step 5: Calculate the remaining amounts of reactants.
Since H₂ is the limiting reagent and we started with 30 L of H₂, we need to find out how much H₂ was consumed:
- According to the stoichiometry, 30 L of H₂ would react to produce 10 L of NH₃:
\[ \text{Volume of H₂ reacted} = \frac{3}{2} \times 10 \text{ L} = 15 \text{ L} \]
Thus, the volume of unused H₂ is:
\[ \text{Unused H₂} = 30 \text{ L} - 15 \text{ L} = 15 \text{ L} \]

For N₂, we started with 30 L. The amount of N₂ that reacted can be calculated as follows:
- 15 L of H₂ reacts with:
\[ \text{Volume of N₂ reacted} = \frac{1}{3} \times 15 \text{ L} = 5 \text{ L} \]
Thus, the volume of unused N₂ is:
\[ \text{Unused N₂} = 30 \text{ L} - 5 \text{ L} = 25 \text{ L} \]

### Step 6: Summarize the final composition of the gaseous mixture.
At the end of the reaction, the composition of the gaseous mixture will be:
- Volume of NH₃ = 10 L
- Volume of unused H₂ = 15 L
- Volume of unused N₂ = 25 L

### Final Answer:
The final composition of the gaseous mixture is:
- 10 L of NH₃
- 15 L of H₂
- 25 L of N₂