56 g of nitrogen and 8 g hydrogen gas are heated in a closed vessel. At equilibrium 34 g of ammnia are present. The equilibrium number of moles of nitrogen, hdregen and ammonia are respectively

Options

A. 1,2,2
B. 2,2,1
C. 1,1,2
D. 2,1,2

Explanation

To solve the problem step by step, we will follow these steps:

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

### Step 2: Calculate the initial number of moles of nitrogen and hydrogen
We will use the formula for calculating the number of moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

- **For Nitrogen (N₂)**:
- Given mass = 56 g
- Molar mass of N₂ = 28 g/mol
\[ \text{Number of moles of } N_2 = \frac{56 \, \text{g}}{28 \, \text{g/mol}} = 2 \, \text{moles} \]

- **For Hydrogen (H₂)**:
- Given mass = 8 g
- Molar mass of H₂ = 2 g/mol
\[ \text{Number of moles of } H_2 = \frac{8 \, \text{g}}{2 \, \text{g/mol}} = 4 \, \text{moles} \]

### Step 3: Set up the initial conditions
Initially, we have:
- Moles of N₂ = 2
- Moles of H₂ = 4
- Moles of NH₃ = 0

### Step 4: Define the change in moles at equilibrium
Let \( x \) be the number of moles of N₂ that react. According to the stoichiometry of the reaction:
- For every 1 mole of N₂ that reacts, 3 moles of H₂ react, and 2 moles of NH₃ are produced.

At equilibrium:
- Moles of N₂ = \( 2 - x \)
- Moles of H₂ = \( 4 - 3x \)
- Moles of NH₃ = \( 2x \)

### Step 5: Use the information given in the problem
We know that at equilibrium, there are 34 g of ammonia (NH₃). First, we need to calculate the number of moles of NH₃:
- Molar mass of NH₃ = 17 g/mol
\[ \text{Number of moles of } NH_3 = \frac{34 \, \text{g}}{17 \, \text{g/mol}} = 2 \, \text{moles} \]

### Step 6: Set up the equation for \( x \)
From our equilibrium expression for NH₃:
\[ 2x = 2 \implies x = 1 \]

### Step 7: Calculate the equilibrium moles of N₂ and H₂
Now we can substitute \( x \) back into our expressions for N₂ and H₂:
- Moles of N₂ at equilibrium:
\[ \text{Moles of } N_2 = 2 - x = 2 - 1 = 1 \]
- Moles of H₂ at equilibrium:
\[ \text{Moles of } H_2 = 4 - 3x = 4 - 3(1) = 4 - 3 = 1 \]

### Step 8: Summarize the results
At equilibrium, we have:
- Moles of N₂ = 1
- Moles of H₂ = 1
- Moles of NH₃ = 2

### Final Answer:
The equilibrium number of moles of nitrogen, hydrogen, and ammonia are respectively:
\[ 1, 1, 2 \]

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