If 100.0 molecules of H, and 39.0 molecules of O, react, how many molecules of H₂O can be produced in the reaction below? 2 H₂(g) + O₂(g) → 2 H₂O(g)
If 100.0 molecules of H, and 39.0 molecules of O, react, how many molecules of H₂O can be produced in the reaction below? 2 H₂(g) + O₂(g) → 2 H₂O(g)
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:**
If 100.0 molecules of \(\text{H}_2\) and 39.0 molecules of \(\text{O}_2\) react, how many molecules of \(\text{H}_2\text{O}\) can be produced in the reaction below?
\[ 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(g) \]
**Explanation:**
To determine the number of \(\text{H}_2\text{O}\) molecules produced, we use the balanced chemical equation:
- According to the equation, 2 molecules of \(\text{H}_2\) react with 1 molecule of \(\text{O}_2\) to produce 2 molecules of \(\text{H}_2\text{O}\).
- With 100.0 molecules of \(\text{H}_2\), you can form a maximum of 50 molecule-pairs with \(\text{O}_2\), producing 100.0 water molecules.
- However, only 39.0 molecules of \(\text{O}_2\) are available, allowing the complete reaction of only 39.0 molecule-pairs, producing 78.0 molecules of \(\text{H}_2\text{O}\).
Hence, the limiting reactant is \(\text{O}_2\) and a total of 78.0 molecules of \(\text{H}_2\text{O}\) can be formed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F96ce4e79-a9b7-4338-8d9e-1cd0f12f5b89%2F82fc26bd-01f0-48d9-aee8-d2138fe1d62a%2Fvz48hta_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
If 100.0 molecules of \(\text{H}_2\) and 39.0 molecules of \(\text{O}_2\) react, how many molecules of \(\text{H}_2\text{O}\) can be produced in the reaction below?
\[ 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(g) \]
**Explanation:**
To determine the number of \(\text{H}_2\text{O}\) molecules produced, we use the balanced chemical equation:
- According to the equation, 2 molecules of \(\text{H}_2\) react with 1 molecule of \(\text{O}_2\) to produce 2 molecules of \(\text{H}_2\text{O}\).
- With 100.0 molecules of \(\text{H}_2\), you can form a maximum of 50 molecule-pairs with \(\text{O}_2\), producing 100.0 water molecules.
- However, only 39.0 molecules of \(\text{O}_2\) are available, allowing the complete reaction of only 39.0 molecule-pairs, producing 78.0 molecules of \(\text{H}_2\text{O}\).
Hence, the limiting reactant is \(\text{O}_2\) and a total of 78.0 molecules of \(\text{H}_2\text{O}\) can be formed.
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