Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![### Reaction Yield Calculation Problem
For the following reaction, 5.69 grams of hydrogen sulfide are mixed with excess oxygen gas. The reaction yields 2.33 grams of water.
\[ \text{2H}_2\text{S (g)} + 3\text{O}_2\text{ (g)} \rightarrow 2\text{H}_2\text{O (l)} + 2\text{SO}_2\text{ (g)} \]
#### Questions:
1. **What is the theoretical yield of water?**
\( \_\_\_\_\_\_\_\_\_\_ \) grams
2. **What is the percent yield for this reaction?**
\( \_\_\_\_\_\_\_\_\_\_ \) %
---
**Explanation:**
- **Theoretical Yield:** This is the maximum amount of product (water in this case) that can be formed from the given amount of reactant (hydrogen sulfide) based on stoichiometric calculations.
- **Percent Yield:** This represents the efficiency of the reaction and is calculated using the formula:
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
To solve these questions, you will need to perform stoichiometric calculations using the molar masses of hydrogen sulfide (H₂S) and water (H₂O) and the given chemical equation.](https://content.bartleby.com/qna-images/question/a527ffdc-16df-4a1f-9953-96956a75d05e/d4b605fb-a9c1-40c9-8521-8e107e0fef5b/zuf8snn.png)
Transcribed Image Text:### Reaction Yield Calculation Problem
For the following reaction, 5.69 grams of hydrogen sulfide are mixed with excess oxygen gas. The reaction yields 2.33 grams of water.
\[ \text{2H}_2\text{S (g)} + 3\text{O}_2\text{ (g)} \rightarrow 2\text{H}_2\text{O (l)} + 2\text{SO}_2\text{ (g)} \]
#### Questions:
1. **What is the theoretical yield of water?**
\( \_\_\_\_\_\_\_\_\_\_ \) grams
2. **What is the percent yield for this reaction?**
\( \_\_\_\_\_\_\_\_\_\_ \) %
---
**Explanation:**
- **Theoretical Yield:** This is the maximum amount of product (water in this case) that can be formed from the given amount of reactant (hydrogen sulfide) based on stoichiometric calculations.
- **Percent Yield:** This represents the efficiency of the reaction and is calculated using the formula:
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]
To solve these questions, you will need to perform stoichiometric calculations using the molar masses of hydrogen sulfide (H₂S) and water (H₂O) and the given chemical equation.
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