Find the weight of `NH_3` in grams when 2.8 kg `N_2` reacts with 1Kg `H_2` ?

Options

A. 6000g
B. 3400g
C. 3000g
D. 5666g

Answer

To solve the problem of finding the weight of ammonia (NH₃) produced when 2.8 kg of nitrogen (N₂) reacts with 1 kg of hydrogen (H₂), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between nitrogen and hydrogen to form ammonia is given by the balanced equation:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Convert the masses of N₂ and H₂ to moles
- **For N₂:**
- Given mass of N₂ = 2.8 kg = 2800 g
- Molar mass of N₂ = 28 g/mol
- Moles of N₂ = \(\frac{2800 \text{ g}}{28 \text{ g/mol}} = 100 \text{ moles}\)

- **For H₂:**
- Given mass of H₂ = 1 kg = 1000 g
- Molar mass of H₂ = 2 g/mol
- Moles of H₂ = \(\frac{1000 \text{ g}}{2 \text{ g/mol}} = 500 \text{ moles}\)

### Step 3: Determine the limiting reagent
From the balanced equation:
- 1 mole of N₂ reacts with 3 moles of H₂.
- Therefore, 100 moles of N₂ would require \(100 \times 3 = 300\) moles of H₂.

Since we have 500 moles of H₂ available, which is more than enough, N₂ is the limiting reagent.

### Step 4: Calculate the moles of NH₃ produced
According to the balanced equation:
- 1 mole of N₂ produces 2 moles of NH₃.
- Therefore, 100 moles of N₂ will produce \(100 \times 2 = 200\) moles of NH₃.

### Step 5: Convert moles of NH₃ to grams
- Molar mass of NH₃ = 14 (N) + 3 (H) = 17 g/mol
- Weight of NH₃ = \(200 \text{ moles} \times 17 \text{ g/mol} = 3400 \text{ g}\)

### Final Answer
The weight of NH₃ produced is **3400 grams**.

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