12. A sample of ammonia (NH,) gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool (Fe is not part of balanced chemical equations). If the total pressure is 866 mm Hg, calculate the partial pressure of N; and H3.
12. A sample of ammonia (NH,) gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool (Fe is not part of balanced chemical equations). If the total pressure is 866 mm Hg, calculate the partial pressure of N; and H3.
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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![### Problem Statement
A sample of ammonia (NH₃) gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool (Fe is not part of balanced chemical equations). If the total pressure is 866 mm Hg, calculate the partial pressure of N₂ and H₂.
### Explanation and Calculation
1. **Balanced Chemical Equation**:
\[
2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2
\]
2. **Total Pressure**:
- Given: Total pressure = 866 mm Hg
3. **Mole Ratio**:
- From the balanced equation, 2 moles of NH₃ yields 1 mole of N₂ and 3 moles of H₂.
- Therefore, the mole ratio of N₂ to H₂ is 1:3.
4. **Partial Pressure Calculations**:
- Let \( P_{N_2} \) be the partial pressure of nitrogen, and \( P_{H_2} \) be the partial pressure of hydrogen.
- According to Dalton’s Law, the total pressure is the sum of the partial pressures:
\[
P_{total} = P_{N_2} + P_{H_2}
\]
- Using the mole ratios:
- If \( x \) is the fraction of the total pressure exerted by N₂, then \( 3x \) is the fraction exerted by H₂, given their ratio.
- Therefore:
\[
P_{N_2} = x, \quad P_{H_2} = 3x
\]
- So:
\[
x + 3x = 866 \text{ mm Hg}
\]
\[
4x = 866
\]
\[
x = 216.5 \text{ mm Hg}
\]
5. **Final Partial Pressures**:
- Partial pressure of \( \text{N}_2 \) (P_{N_2}) = 216.5 mm Hg
- Partial pressure of \( \text{H}_2 \) (P_{H_2}) = 649.5 mm Hg (since \( 3x = 649.5 \))
In conclusion, the partial pressure of nitrogen](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2bde565e-d7c1-4791-8913-55a421a09e09%2F82a0869d-c5c4-4967-8646-11a8fa2c91c3%2F6p4o1o7_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A sample of ammonia (NH₃) gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool (Fe is not part of balanced chemical equations). If the total pressure is 866 mm Hg, calculate the partial pressure of N₂ and H₂.
### Explanation and Calculation
1. **Balanced Chemical Equation**:
\[
2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2
\]
2. **Total Pressure**:
- Given: Total pressure = 866 mm Hg
3. **Mole Ratio**:
- From the balanced equation, 2 moles of NH₃ yields 1 mole of N₂ and 3 moles of H₂.
- Therefore, the mole ratio of N₂ to H₂ is 1:3.
4. **Partial Pressure Calculations**:
- Let \( P_{N_2} \) be the partial pressure of nitrogen, and \( P_{H_2} \) be the partial pressure of hydrogen.
- According to Dalton’s Law, the total pressure is the sum of the partial pressures:
\[
P_{total} = P_{N_2} + P_{H_2}
\]
- Using the mole ratios:
- If \( x \) is the fraction of the total pressure exerted by N₂, then \( 3x \) is the fraction exerted by H₂, given their ratio.
- Therefore:
\[
P_{N_2} = x, \quad P_{H_2} = 3x
\]
- So:
\[
x + 3x = 866 \text{ mm Hg}
\]
\[
4x = 866
\]
\[
x = 216.5 \text{ mm Hg}
\]
5. **Final Partial Pressures**:
- Partial pressure of \( \text{N}_2 \) (P_{N_2}) = 216.5 mm Hg
- Partial pressure of \( \text{H}_2 \) (P_{H_2}) = 649.5 mm Hg (since \( 3x = 649.5 \))
In conclusion, the partial pressure of nitrogen
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