20.0 kg of `N_(2)(g)` and 3.0 kg of `H_(2)(g)` are mixed to produce `NH_(3)(g)`. The amount of `NH_(3)(g)` formed is

Options

A. 17 kg
B. 34 kg
C. 20 kg
D. 3 kg

Explanation

To solve the problem of how much ammonia (NH₃) can be produced from 20.0 kg of nitrogen gas (N₂) and 3.0 kg of hydrogen gas (H₂), we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the formation of ammonia is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

### Step 2: Calculate the Molar Masses
- Molar mass of \( N_2 \) = 28 g/mol (since N = 14 g/mol)
- Molar mass of \( H_2 \) = 2 g/mol (since H = 1 g/mol)
- Molar mass of \( NH_3 \) = 17 g/mol (N = 14 g/mol + 3H = 3 g/mol)

### Step 3: Convert Mass to Moles
Convert the given masses of nitrogen and hydrogen to moles:
- For nitrogen:
\[ \text{Moles of } N_2 = \frac{20,000 \text{ g}}{28 \text{ g/mol}} \approx 714.29 \text{ mol} \]
- For hydrogen:
\[ \text{Moles of } H_2 = \frac{3,000 \text{ g}}{2 \text{ g/mol}} = 1500 \text{ mol} \]

### Step 4: Determine the Limiting Reagent
From the balanced equation, we know that 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \). Therefore:
- Moles of \( H_2 \) required for 714.29 moles of \( N_2 \):
\[ 714.29 \text{ mol } N_2 \times 3 = 2142.87 \text{ mol } H_2 \]
Since we only have 1500 moles of \( H_2 \), hydrogen is the limiting reagent.

### Step 5: Calculate the Amount of Ammonia Produced
Using the limiting reagent (H₂), we can calculate the amount of ammonia produced:
From the balanced equation, 3 moles of \( H_2 \) produce 2 moles of \( NH_3 \):
- Moles of \( NH_3 \) produced from 1500 moles of \( H_2 \):
\[ \text{Moles of } NH_3 = \frac{2}{3} \times 1500 \approx 1000 \text{ mol} \]

### Step 6: Convert Moles of Ammonia to Mass
Convert the moles of ammonia produced to mass:
\[ \text{Mass of } NH_3 = 1000 \text{ mol} \times 17 \text{ g/mol} = 17000 \text{ g} = 17.0 \text{ kg} \]

### Final Answer
The amount of ammonia (NH₃) formed is **17.0 kg**.

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