A mixture of `N_2` and `H_2` is caused to react in a closed container to form `NH_3`. The reaction decreases before any of the reactant has been totally consumed. At this stage, 2 moles each of `N_2`, `H_2 `and` NH_3` are present. Then the weight of `N_2` and `H_2` 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 step by step, we will analyze the reaction between nitrogen and hydrogen to form ammonia, and calculate the original weights of nitrogen and hydrogen.

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Define the initial moles of reactants
Let the initial moles of nitrogen (\(N_2\)) be \(A\) and the initial moles of hydrogen (\(H_2\)) be \(B\).

### Step 3: Determine the change in moles at equilibrium
At equilibrium, we know that there are 2 moles of \(N_2\), 2 moles of \(H_2\), and 2 moles of \(NH_3\).

Using the stoichiometry of the reaction:
- For every 1 mole of \(N_2\) that reacts, 3 moles of \(H_2\) are consumed, and 2 moles of \(NH_3\) are produced.

Let \(x\) be the number of moles of \(N_2\) that reacted. Thus:
- Moles of \(N_2\) at equilibrium = \(A - x\)
- Moles of \(H_2\) at equilibrium = \(B - 3x\)
- Moles of \(NH_3\) at equilibrium = \(2x\)

### Step 4: Set up equations based on equilibrium conditions
From the problem, we know:
1. \(A - x = 2\) (for \(N_2\))
2. \(B - 3x = 2\) (for \(H_2\))
3. \(2x = 2\) (for \(NH_3\))

From the third equation, we can solve for \(x\):
\[ 2x = 2 \implies x = 1 \]

### Step 5: Substitute \(x\) back into the equations
Now substituting \(x = 1\) into the first two equations:
1. \(A - 1 = 2 \implies A = 3\)
2. \(B - 3(1) = 2 \implies B - 3 = 2 \implies B = 5\)

### Step 6: Calculate the weights of \(N_2\) and \(H_2\)
Now that we have the initial moles:
- Moles of \(N_2 = 3\)
- Moles of \(H_2 = 5\)

Next, we calculate the weights using their molar masses:
- Molar mass of \(N_2 = 28 \, \text{g/mol}\)
- Molar mass of \(H_2 = 2 \, \text{g/mol}\)

Calculating the weights:
- Weight of \(N_2 = 3 \, \text{moles} \times 28 \, \text{g/mol} = 84 \, \text{g}\)
- Weight of \(H_2 = 5 \, \text{moles} \times 2 \, \text{g/mol} = 10 \, \text{g}\)

### Final Answer
The original weights of \(N_2\) and \(H_2\) were 84 grams and 10 grams, respectively.

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