Question

Nitrogen gas and hydrogen gas react to produce ammonia according to the following equation.

N2+3H2→2NH3

The atomic mass of nitrogen is 14 g/mol
. The atomic mass of hydrogen is 1 g/mol
. When the reaction produces 68
grams of ammonia, how many grams of nitrogen were present initially?

(1 point)
Responses

12
12

56
56

112
112

4

Answer 1

To find out how many grams of nitrogen were present initially when the reaction produced 68 grams of ammonia (NH₃), we can follow these steps:

1. **Calculate the molar mass of ammonia (NH₃)**:
- Nitrogen (N): 14 g/mol
- Hydrogen (H): 1 g/mol
- Therefore, the molar mass of NH₃ = 14 g/mol + (3 × 1 g/mol) = 14 g/mol + 3 g/mol = 17 g/mol.

2. **Determine the number of moles of ammonia produced**:
- Mass of NH₃ = 68 g
- Moles of NH₃ = mass / molar mass = 68 g / 17 g/mol = 4 moles of NH₃.

3. **Use the stoichiometry of the reaction**:
- According to the balanced equation: N₂ + 3H₂ → 2NH₃,
- It indicates that 2 moles of NH₃ are produced from 1 mole of N₂.
- Therefore, from 4 moles of NH₃: (4 moles NH₃) × (1 mole N₂ / 2 moles NH₃) = 2 moles of N₂.

4. **Calculate the mass of nitrogen (N₂)**:
- Molar mass of N₂ = 2 × 14 g/mol = 28 g/mol.
- Mass of N₂ = moles × molar mass = 2 moles × 28 g/mol = 56 g.

Thus, when 68 grams of ammonia are produced, there were initially **56 grams of nitrogen** present. The correct response is **56**.