28g of `N_2` and 6g of `H_2` were mixed. At equilibrium 17g of `NH_3` was formed. The weight of `N_2 and H_2` at equilibrium are respectively

Options

A. 11 g & zero
B. 19 & 3g
C. 14g & 3g
D. 11g & 3g

Answer

To solve the problem, we need to determine the weights of nitrogen (N₂) and hydrogen (H₂) at equilibrium after mixing 28 g of N₂ and 6 g of H₂, resulting in the formation of 17 g of ammonia (NH₃). We will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the formation of ammonia from nitrogen and hydrogen is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step 2: Calculate the initial moles of N₂ and H₂
- **Molar mass of N₂**:
\[ \text{Molar mass of N₂} = 2 \times 14 = 28 \, \text{g/mol} \]
- **Moles of N₂**:
\[ \text{Moles of N₂} = \frac{\text{Given weight}}{\text{Molar mass}} = \frac{28 \, \text{g}}{28 \, \text{g/mol}} = 1 \, \text{mol} \]

- **Molar mass of H₂**:
\[ \text{Molar mass of H₂} = 2 \times 1 = 2 \, \text{g/mol} \]
- **Moles of H₂**:
\[ \text{Moles of H₂} = \frac{6 \, \text{g}}{2 \, \text{g/mol}} = 3 \, \text{mol} \]

### Step 3: Determine the moles of NH₃ formed
- **Molar mass of NH₃**:
\[ \text{Molar mass of NH₃} = 14 + (3 \times 1) = 17 \, \text{g/mol} \]
- **Moles of NH₃ formed**:
\[ \text{Moles of NH₃} = \frac{17 \, \text{g}}{17 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 4: Relate moles of reactants to moles of products
From the balanced equation:
- 1 mole of N₂ produces 2 moles of NH₃.
- Therefore, 1 mole of NH₃ requires \( \frac{1}{2} \) moles of N₂ and \( \frac{3}{2} \) moles of H₂.

### Step 5: Calculate the change in moles
Let \( x \) be the moles of N₂ that reacted:
- For NH₃: \( 2x = 1 \) (since 1 mole of NH₃ is formed)
- Thus, \( x = \frac{1}{2} \) moles of N₂ reacted.

### Step 6: Calculate remaining moles of N₂ and H₂
- **Remaining moles of N₂**:
\[ \text{Remaining moles of N₂} = 1 - \frac{1}{2} = \frac{1}{2} \, \text{mol} \]
- **Moles of H₂ reacted**:
\[ \text{Moles of H₂ reacted} = 3 \times \frac{1}{2} = \frac{3}{2} \, \text{mol} \]
- **Remaining moles of H₂**:
\[ \text{Remaining moles of H₂} = 3 - \frac{3}{2} = \frac{3}{2} \, \text{mol} \]

### Step 7: Calculate the weights of N₂ and H₂ at equilibrium
- **Weight of N₂ at equilibrium**:
\[ \text{Weight of N₂} = \text{Moles of N₂} \times \text{Molar mass of N₂} = \frac{1}{2} \times 28 = 14 \, \text{g} \]
- **Weight of H₂ at equilibrium**:
\[ \text{Weight of H₂} = \text{Moles of H₂} \times \text{Molar mass of H₂} = \frac{3}{2} \times 2 = 3 \, \text{g} \]

### Final Answer
The weights of N₂ and H₂ at equilibrium are:
- Weight of N₂ = 14 g
- Weight of H₂ = 3 g