A mixture of N2 and H2 is caused to react in a closed container to form NH3. The reaction ceases before any of the reactant has been totally consumed. At this stage, 2 moles each of N2, H2 and NH3 are present. Then the weight of N2 and H2 present originally were respectively

Options

A. 112 g and 8 g
B. 84 g and 10 g
C. 84 g and 8 g
D. 122 g and 10 g

Answer

To solve the problem, we need to analyze the reaction between nitrogen (N2) and hydrogen (H2) to form ammonia (NH3) and determine the initial amounts of N2 and H2 based on the equilibrium conditions provided.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

2. **Define Initial Moles:**
Let the initial moles of N2 be \( a \) and the initial moles of H2 be \( b \).

3. **Define Change in Moles:**
Let \( x \) be the moles of N2 that reacted. According to the stoichiometry of the reaction:
- Moles of N2 remaining = \( a - x \)
- Moles of H2 remaining = \( b - 3x \)
- Moles of NH3 formed = \( 2x \)

4. **Set Up Equilibrium Conditions:**
At equilibrium, we are given that there are 2 moles of each substance:
\[ a - x = 2 \quad (1) \]
\[ b - 3x = 2 \quad (2) \]
\[ 2x = 2 \quad (3) \]

5. **Solve for \( x \):**
From equation (3):
\[ 2x = 2 \implies x = 1 \]

6. **Substitute \( x \) into Equations (1) and (2):**
Substitute \( x = 1 \) into equation (1):
\[ a - 1 = 2 \implies a = 3 \]
Substitute \( x = 1 \) into equation (2):
\[ b - 3(1) = 2 \implies b - 3 = 2 \implies b = 5 \]

7. **Determine Initial Moles:**
- Initial moles of N2 = \( a = 3 \)
- Initial moles of H2 = \( b = 5 \)

8. **Calculate the Mass of Each Gas:**
- Molar mass of N2 = 28 g/mol
- Molar mass of H2 = 2 g/mol

For N2:
\[ \text{Mass of N2} = \text{moles} \times \text{molar mass} = 3 \times 28 = 84 \text{ g} \]

For H2:
\[ \text{Mass of H2} = \text{moles} \times \text{molar mass} = 5 \times 2 = 10 \text{ g} \]

9. **Final Answer:**
The initial weights of N2 and H2 are:
- Weight of N2 = 84 g
- Weight of H2 = 10 g