10. In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: C6H1206 (s) →2C2H5OH + 2CO2 (9) If 4.26 g of glucose are reacted and 1.14 L of CO2 gas are collected at 306 K and 0.985 atm. what is the percent yield of the reaction?
10. In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: C6H1206 (s) →2C2H5OH + 2CO2 (9) If 4.26 g of glucose are reacted and 1.14 L of CO2 gas are collected at 306 K and 0.985 atm. what is the percent yield of the reaction?
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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![### Educational Content on Alcohol Fermentation and Yield Calculation
**Alcohol Fermentation**
In the process of alcohol fermentation, yeast converts glucose into ethanol and carbon dioxide. The chemical equation for this reaction is:
\[ \text{C}_6\text{H}_{12}\text{O}_6 \,(\text{s}) \rightarrow 2\text{C}_2\text{H}_5\text{OH} \,(\ell) + 2\text{CO}_2 \,(\text{g}) \]
**Problem Statement**
If 4.26 grams of glucose are reacted and 1.14 liters of \(\text{CO}_2\) gas is collected at 306 K and 0.985 atm, what is the percent yield of the reaction?
**Explanation of the Diagram**
The image contains handwritten calculations related to the given problem. The calculations demonstrate how to determine the number of moles of \(\text{CO}_2\) produced using the ideal gas law.
1. **Ideal Gas Law Calculation**:
\[
PV = nRT
\]
- \(P\) = 719.9 mm Hg (converted to atm as appropriate),
- \(V\) = 0.529 L (as noted in the handwritten text),
- \(R\) (the ideal gas constant),
- \(T\) = 296 K.
2. **Substituted Values**:
- \(\frac{719.9 \, \text{mm Hg} \times 0.529 \, \text{L}}{62.36 \, \text{mm Hg} \cdot \text{L/mol} \cdot \text{K} \times 296 \, \text{K}}\)
- The calculation results in 0.0128 moles of \(\text{CO}_2\).
3. **Percent Yield Calculation**:
- Percent yield can be calculated using the actual yield and theoretical yield:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]
**Note**: The theoretical yield must be calculated from the stoichiometry of the reaction, considering 4.26 g of glucose used.
This explanation aims to guide students through the process of understanding alcohol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7de42a67-181a-4f2f-af21-0f59713bb114%2Fa1a37354-d607-4228-b7cb-acfe403d448c%2Fk94zaz.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Content on Alcohol Fermentation and Yield Calculation
**Alcohol Fermentation**
In the process of alcohol fermentation, yeast converts glucose into ethanol and carbon dioxide. The chemical equation for this reaction is:
\[ \text{C}_6\text{H}_{12}\text{O}_6 \,(\text{s}) \rightarrow 2\text{C}_2\text{H}_5\text{OH} \,(\ell) + 2\text{CO}_2 \,(\text{g}) \]
**Problem Statement**
If 4.26 grams of glucose are reacted and 1.14 liters of \(\text{CO}_2\) gas is collected at 306 K and 0.985 atm, what is the percent yield of the reaction?
**Explanation of the Diagram**
The image contains handwritten calculations related to the given problem. The calculations demonstrate how to determine the number of moles of \(\text{CO}_2\) produced using the ideal gas law.
1. **Ideal Gas Law Calculation**:
\[
PV = nRT
\]
- \(P\) = 719.9 mm Hg (converted to atm as appropriate),
- \(V\) = 0.529 L (as noted in the handwritten text),
- \(R\) (the ideal gas constant),
- \(T\) = 296 K.
2. **Substituted Values**:
- \(\frac{719.9 \, \text{mm Hg} \times 0.529 \, \text{L}}{62.36 \, \text{mm Hg} \cdot \text{L/mol} \cdot \text{K} \times 296 \, \text{K}}\)
- The calculation results in 0.0128 moles of \(\text{CO}_2\).
3. **Percent Yield Calculation**:
- Percent yield can be calculated using the actual yield and theoretical yield:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]
**Note**: The theoretical yield must be calculated from the stoichiometry of the reaction, considering 4.26 g of glucose used.
This explanation aims to guide students through the process of understanding alcohol
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