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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![**Question:**
How many liters of ammonia (NH₃) are produced when 4.7 L of nitrogen gas reacts with excess hydrogen at STP?
**Explanation:**
This question involves a chemical reaction under standard temperature and pressure (STP) conditions. At STP, 1 mole of any gas occupies 22.4 liters. The reaction of nitrogen gas (N₂) with hydrogen gas (H₂) to form ammonia is represented by the balanced chemical equation:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
From the equation, one mole of nitrogen gas produces two moles of ammonia. Therefore, 22.4 liters of nitrogen gas would theoretically produce 44.8 liters of ammonia at STP.
For 4.7 liters of nitrogen gas:
\[ \text{Volume of } NH_3 = \left(\frac{2 \times 4.7 \, \text{L}}{1}\right) = 9.4 \, \text{L of } NH_3 \]
Thus, 9.4 liters of ammonia are produced under these conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb077f2ab-172b-438d-8d7b-87f013063300%2Fcc6cef82-bb7f-4222-9b48-aa46e33edb0e%2Fpx1uyl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
How many liters of ammonia (NH₃) are produced when 4.7 L of nitrogen gas reacts with excess hydrogen at STP?
**Explanation:**
This question involves a chemical reaction under standard temperature and pressure (STP) conditions. At STP, 1 mole of any gas occupies 22.4 liters. The reaction of nitrogen gas (N₂) with hydrogen gas (H₂) to form ammonia is represented by the balanced chemical equation:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
From the equation, one mole of nitrogen gas produces two moles of ammonia. Therefore, 22.4 liters of nitrogen gas would theoretically produce 44.8 liters of ammonia at STP.
For 4.7 liters of nitrogen gas:
\[ \text{Volume of } NH_3 = \left(\frac{2 \times 4.7 \, \text{L}}{1}\right) = 9.4 \, \text{L of } NH_3 \]
Thus, 9.4 liters of ammonia are produced under these conditions.
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