Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N2 (g) + 3H2(g) → 2NH3 (g) Determine the mass of N2 (g) required to completely react with 18.5 g of H2.
Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N2 (g) + 3H2(g) → 2NH3 (g) Determine the mass of N2 (g) required to completely react with 18.5 g of H2.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![## Determining the Mass of Nitrogen Required for a Reaction
**Problem Statement:**
Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia:
\[ \text{N}_2 (\text{g}) + 3\text{H}_2 (\text{g}) \rightarrow 2\text{NH}_3 (\text{g}) \]
Determine the mass of nitrogen (N\(_2\)) required to completely react with 18.5 grams of hydrogen (H\(_2\)).
### Solution:
1. Write the balanced chemical equation, which is already provided:
\[ \text{N}_2 (\text{g}) + 3\text{H}_2 (\text{g}) \rightarrow 2\text{NH}_3 (\text{g}) \]
2. Convert the mass of hydrogen gas (H\(_2\)) to moles using its molar mass:
- Molar mass of H\(_2\) = 2 g/mol
- Moles of H\(_2\):
\[ \text{Moles of H}_2 = \frac{18.5 \text{ g}}{2 \text{ g/mol}} = 9.25 \text{ mol} \]
3. Use the stoichiometry of the balanced equation to relate moles of H\(_2\) to moles of N\(_2\):
- According to the balanced equation, 3 moles of H\(_2\) react with 1 mole of N\(_2\).
- Moles of N\(_2\) required:
\[ \text{Moles of N}_2 = \frac{9.25 \text{ mol H}_2}{3} = 3.083 \text{ mol N}_2 \]
4. Convert moles of nitrogen gas (N\(_2\)) to mass using its molar mass:
- Molar mass of N\(_2\) = 28 g/mol
- Mass of N\(_2\) required:
\[ \text{Mass of N}_2 = 3.083 \text{ mol} \times 28 \text{ g/mol} = 86.324 \text{ g} \]
### Conclusion:
To completely react with 18.5 grams of hydrogen gas (H\(_2\)), 86.324 grams](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35780925-5e45-4c26-92de-ecbbabfbfb48%2F2747c902-1434-403c-8bdc-6aeff6392e4f%2Fdubx3pm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Determining the Mass of Nitrogen Required for a Reaction
**Problem Statement:**
Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia:
\[ \text{N}_2 (\text{g}) + 3\text{H}_2 (\text{g}) \rightarrow 2\text{NH}_3 (\text{g}) \]
Determine the mass of nitrogen (N\(_2\)) required to completely react with 18.5 grams of hydrogen (H\(_2\)).
### Solution:
1. Write the balanced chemical equation, which is already provided:
\[ \text{N}_2 (\text{g}) + 3\text{H}_2 (\text{g}) \rightarrow 2\text{NH}_3 (\text{g}) \]
2. Convert the mass of hydrogen gas (H\(_2\)) to moles using its molar mass:
- Molar mass of H\(_2\) = 2 g/mol
- Moles of H\(_2\):
\[ \text{Moles of H}_2 = \frac{18.5 \text{ g}}{2 \text{ g/mol}} = 9.25 \text{ mol} \]
3. Use the stoichiometry of the balanced equation to relate moles of H\(_2\) to moles of N\(_2\):
- According to the balanced equation, 3 moles of H\(_2\) react with 1 mole of N\(_2\).
- Moles of N\(_2\) required:
\[ \text{Moles of N}_2 = \frac{9.25 \text{ mol H}_2}{3} = 3.083 \text{ mol N}_2 \]
4. Convert moles of nitrogen gas (N\(_2\)) to mass using its molar mass:
- Molar mass of N\(_2\) = 28 g/mol
- Mass of N\(_2\) required:
\[ \text{Mass of N}_2 = 3.083 \text{ mol} \times 28 \text{ g/mol} = 86.324 \text{ g} \]
### Conclusion:
To completely react with 18.5 grams of hydrogen gas (H\(_2\)), 86.324 grams
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