Question

Nitrogen gas and hydrogen gas react to produce ammonia according to the following equation.
The atomic mass of nitrogen is . The atomic mass of hydrogen is . When the reaction
produces grams of ammonia, how many grams of nitrogen were present initially?
(1 point)
N2 + 3H2 → 2NH3 N2 + 3H2 → 2NH3
14 g/mol 14 g / mol 1 g/mol 1 g / mol
6868
1212
112112
5656
4

Answer 1

To determine how many grams of nitrogen were present initially when producing a certain amount of ammonia, we can follow these steps:

1. **Write down the balanced chemical equation**:
\[
N_2 + 3H_2 \rightarrow 2NH_3
\]

2. **Determine the molar masses**:
- Molar mass of \(N_2\) (for nitrogen) = \(2 \times 14 \, \text{g/mol} = 28 \, \text{g/mol}\)
- Molar mass of \(H_2\) (for hydrogen) = \(2 \times 1 \, \text{g/mol} = 2 \, \text{g/mol}\)
- Molar mass of \(NH_3\) (for ammonia) = \(14 \, \text{g/mol} + 3 \times 1 \, \text{g/mol} = 17 \, \text{g/mol}\)

3. **Examine the stoichiometry**:
According to the balanced equation:
- 1 mole of \(N_2\) produces 2 moles of \(NH_3\).
- 1 mole of \(N_2\) corresponds to \(28 \, \text{g}\) of nitrogen and produces \(34 \, \text{g}\) of \(NH_3\) (since \(2 \times 17 \, \text{g/mol} = 34 \, \text{g}\)).

4. **Calculate the mass of nitrogen based on the mass of ammonia produced**:
If \(x\) grams of ammonia is produced, we can set up a proportion based on the stoichiometric ratios:
\[
\frac{28 \, \text{g} \, N_2}{34 \, \text{g} \, NH_3} = \frac{m \, \text{g} \, N_2}{x \, \text{g} \, NH_3}
\]
Rearranging this gives:
\[
m = \frac{28 \, \text{g}}{34 \, \text{g}} \cdot x
\]

5. **Calculate the amount of nitrogen for the specific amount of ammonia**:
- If \(x = 68 \, \text{g}\) (assuming the amount of ammonia produced is given as 68 g), we can substitute \(x\):
\[
m = \frac{28}{34} \cdot 68
\]
\[
m = \frac{28 \times 68}{34} = 56 \, \text{grams of } N_2
\]

Thus, if 68 grams of ammonia were produced, there were initially **56 grams of nitrogen** present.