28 g of `N_(2)` and 6 g of `H_(2)` were mixed. At equilibrium 17 g `NH_(3)` was produced. The weight of `N_(2)` and `H_(2)` at equilibrium are respectively

Options

A. 11 g, zero
B. 1 g, 3 g
C. 14 g, 3 g
D. 11 g, 3 g

Answer

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

### Step 1: Write the balanced chemical equation
The balanced equation for the formation of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) is:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]

### Step 2: Calculate the moles of NH₃ produced
We know that 17 g of NH₃ is produced. The molar mass of NH₃ is 17 g/mol. Therefore, the number of moles of NH₃ produced is:
\[ \text{Moles of NH}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{17 \text{ g}}{17 \text{ g/mol}} = 1 \text{ mole} \]

### Step 3: Determine the moles of N₂ and H₂ used
From the balanced equation, we see that 1 mole of N₂ produces 2 moles of NH₃. Therefore, to produce 1 mole of NH₃, we need:
- Moles of N₂ used = \( \frac{1}{2} \) moles
- Moles of H₂ used = \( \frac{3}{2} \) moles

### Step 4: Calculate the mass of N₂ and H₂ used
Now, we can calculate the mass of N₂ and H₂ that were used:
- Molar mass of N₂ = 28 g/mol
- Molar mass of H₂ = 2 g/mol

**Mass of N₂ used:**
\[ \text{Mass of N}_2 = \text{moles} \times \text{molar mass} = \frac{1}{2} \text{ moles} \times 28 \text{ g/mol} = 14 \text{ g} \]

**Mass of H₂ used:**
\[ \text{Mass of H}_2 = \text{moles} \times \text{molar mass} = \frac{3}{2} \text{ moles} \times 2 \text{ g/mol} = 3 \text{ g} \]

### Step 5: Calculate the remaining mass of N₂ and H₂ at equilibrium
Initially, we had:
- Mass of N₂ = 28 g
- Mass of H₂ = 6 g

Now, we subtract the masses used from the initial masses:
- Remaining mass of N₂:
\[ \text{Remaining N}_2 = 28 \text{ g} - 14 \text{ g} = 14 \text{ g} \]

- Remaining mass of H₂:
\[ \text{Remaining H}_2 = 6 \text{ g} - 3 \text{ g} = 3 \text{ g} \]

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

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