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

Answer

To find 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 balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Calculate the molar masses
- Molar mass of \( N_2 \) (Nitrogen gas) = 28 g/mol (since each nitrogen atom has a mass of 14 g)
- Molar mass of \( H_2 \) (Hydrogen gas) = 2 g/mol (since each hydrogen atom has a mass of 1 g)
- Molar mass of \( NH_3 \) (Ammonia) = 17 g/mol (14 g from nitrogen and 3 g from hydrogen)

### Step 3: Convert the given masses to grams
- Given mass of \( N_2 = 2.8 \, \text{kg} = 2800 \, \text{g} \)
- Given mass of \( H_2 = 1 \, \text{kg} = 1000 \, \text{g} \)

### Step 4: Calculate the number of moles of \( N_2 \) and \( H_2 \)
- Moles of \( N_2 \):
\[ \text{Moles of } N_2 = \frac{2800 \, \text{g}}{28 \, \text{g/mol}} = 100 \, \text{moles} \]
- Moles of \( H_2 \):
\[ \text{Moles of } H_2 = \frac{1000 \, \text{g}}{2 \, \text{g/mol}} = 500 \, \text{moles} \]

### Step 5: Determine the limiting reactant
From the balanced equation, 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \). Therefore, for 100 moles of \( N_2 \), we need:
\[ 100 \, \text{moles of } N_2 \times 3 = 300 \, \text{moles of } H_2 \]
Since we have 500 moles of \( H_2 \), which is more than enough, \( N_2 \) is the limiting reactant.

### Step 6: Calculate the moles of \( NH_3 \) produced
According to the balanced equation, 1 mole of \( N_2 \) produces 2 moles of \( NH_3 \). Therefore, 100 moles of \( N_2 \) will produce:
\[ 100 \, \text{moles of } N_2 \times 2 = 200 \, \text{moles of } NH_3 \]

### Step 7: Calculate the weight of \( NH_3 \) produced
Now, we can calculate the weight of \( NH_3 \):
\[ \text{Weight of } NH_3 = \text{Moles of } NH_3 \times \text{Molar mass of } NH_3 \]
\[ = 200 \, \text{moles} \times 17 \, \text{g/mol} = 3400 \, \text{g} \]

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

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