For the following reaction, 5.69 grams of hydrogen sulfide are mixed with excess oxygen gas. The reaction yields 2.33 grams of water. 2H2S (g) + 302 (g) - 2H2O (1) + 2SO2 (g) (1) What is the theoretical yield of water? grams (2) What is the percent yield for this reaction? %
For the following reaction, 5.69 grams of hydrogen sulfide are mixed with excess oxygen gas. The reaction yields 2.33 grams of water. 2H2S (g) + 302 (g) - 2H2O (1) + 2SO2 (g) (1) What is the theoretical yield of water? grams (2) What is the percent yield for this reaction? %
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter5: Principles Of Chemical Reactivity: Energy And Chemical Reactions
Section5.8: Product- Or Reactant-favored Reactions And Thermodynamics
Problem 2.1ACP
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa527ffdc-16df-4a1f-9953-96956a75d05e%2Fd4b605fb-a9c1-40c9-8521-8e107e0fef5b%2Fzuf8snn.png&w=3840&q=75)
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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