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
112
112
56
56
4
4
12
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
112
112
56
56
4
4
12
Answers
Answered by
GPT-4o mini
To determine how many grams of nitrogen were present initially when 68 grams of ammonia (NH₃) were produced, we can use stoichiometry based on the balanced chemical reaction:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]
From the equation, we can see that 2 moles of ammonia (NH₃) are produced from 1 mole of nitrogen (N₂). We will follow these steps:
1. **Calculate the molar mass of ammonia (NH₃)**:
- Nitrogen (N) has an atomic mass of 14 g/mol.
- Hydrogen (H) has an atomic mass of 1 g/mol.
- Therefore, the molar mass of NH₃ is:
\[
14 \, \text{g/mol (N)} + 3 \times 1 \, \text{g/mol (H)} = 14 + 3 = 17 \, \text{g/mol}
\]
2. **Determine the moles of ammonia produced**:
- Given that 68 grams of NH₃ were produced:
\[
\text{Moles of NH₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{68 \, \text{g}}{17 \, \text{g/mol}} = 4 \, \text{moles}
\]
3. **Use stoichiometry to find moles of nitrogen used**:
- The reaction shows that 2 moles of NH₃ are produced from 1 mole of N₂, so for every 2 moles of NH₃, we consume 1 mole of N₂:
\[
\text{Moles of N₂} = \frac{4 \, \text{moles NH₃}}{2} = 2 \, \text{moles N₂}
\]
4. **Calculate the mass of nitrogen used**:
- Molar mass of nitrogen (N₂) is:
\[
2 \times 14 \, \text{g/mol} = 28 \, \text{g/mol}
\]
- The mass of nitrogen used is:
\[
\text{mass of N₂} = \text{moles} \times \text{molar mass} = 2 \, \text{moles} \times 28 \, \text{g/mol} = 56 \, \text{grams}
\]
Therefore, the amount of nitrogen present initially was **56 grams**.
The correct answer is:
**56**
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]
From the equation, we can see that 2 moles of ammonia (NH₃) are produced from 1 mole of nitrogen (N₂). We will follow these steps:
1. **Calculate the molar mass of ammonia (NH₃)**:
- Nitrogen (N) has an atomic mass of 14 g/mol.
- Hydrogen (H) has an atomic mass of 1 g/mol.
- Therefore, the molar mass of NH₃ is:
\[
14 \, \text{g/mol (N)} + 3 \times 1 \, \text{g/mol (H)} = 14 + 3 = 17 \, \text{g/mol}
\]
2. **Determine the moles of ammonia produced**:
- Given that 68 grams of NH₃ were produced:
\[
\text{Moles of NH₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{68 \, \text{g}}{17 \, \text{g/mol}} = 4 \, \text{moles}
\]
3. **Use stoichiometry to find moles of nitrogen used**:
- The reaction shows that 2 moles of NH₃ are produced from 1 mole of N₂, so for every 2 moles of NH₃, we consume 1 mole of N₂:
\[
\text{Moles of N₂} = \frac{4 \, \text{moles NH₃}}{2} = 2 \, \text{moles N₂}
\]
4. **Calculate the mass of nitrogen used**:
- Molar mass of nitrogen (N₂) is:
\[
2 \times 14 \, \text{g/mol} = 28 \, \text{g/mol}
\]
- The mass of nitrogen used is:
\[
\text{mass of N₂} = \text{moles} \times \text{molar mass} = 2 \, \text{moles} \times 28 \, \text{g/mol} = 56 \, \text{grams}
\]
Therefore, the amount of nitrogen present initially was **56 grams**.
The correct answer is:
**56**