When copper is heated with an excess of sulfur, copper (1) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 6.36 g copper (1) sulfide. What is the percent yield?
When copper is heated with an excess of sulfur, copper (1) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 6.36 g copper (1) sulfide. What is the percent yield?
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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![**Chemical Reaction and Percent Yield Calculation**
**Scenario:**
When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 6.36 grams of copper(I) sulfide. The question to consider is: What is the percent yield of this reaction?
**Explanation:**
In this chemical reaction, we start with a known quantity of copper and an excess of sulfur. The goal is to form copper(I) sulfide, and we need to determine how efficient the process was, given the actual yield of the product.
**Percent Yield Calculation:**
1. **Determine Theoretical Yield:**
- Use stoichiometry to calculate the theoretical yield of copper(I) sulfide based on 0.0970 moles of copper.
2. **Calculate Percent Yield:**
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%
\]
The provided data indicates that the actual yield is 6.36 grams. This information will be used to find the percent yield by comparing it against the calculated theoretical yield.
Understanding this concept is essential in evaluating the efficiency of chemical reactions in real-world scenarios.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65936c84-20b2-4046-a438-4a3445f2d7a5%2Ff311a857-074b-4457-a916-ba4d66f32e3b%2Fg28kwkb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Reaction and Percent Yield Calculation**
**Scenario:**
When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 6.36 grams of copper(I) sulfide. The question to consider is: What is the percent yield of this reaction?
**Explanation:**
In this chemical reaction, we start with a known quantity of copper and an excess of sulfur. The goal is to form copper(I) sulfide, and we need to determine how efficient the process was, given the actual yield of the product.
**Percent Yield Calculation:**
1. **Determine Theoretical Yield:**
- Use stoichiometry to calculate the theoretical yield of copper(I) sulfide based on 0.0970 moles of copper.
2. **Calculate Percent Yield:**
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%
\]
The provided data indicates that the actual yield is 6.36 grams. This information will be used to find the percent yield by comparing it against the calculated theoretical yield.
Understanding this concept is essential in evaluating the efficiency of chemical reactions in real-world scenarios.
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