Give the percent yield when 28.16 g of CO_2[/math} are formed from the reaction of 4.000 moles of [math]C_8H_18 with 8.000 moles of 02. 2C3H18+ 2502→16CO2 + 18H2O
Give the percent yield when 28.16 g of CO_2[/math} are formed from the reaction of 4.000 moles of [math]C_8H_18 with 8.000 moles of 02. 2C3H18+ 2502→16CO2 + 18H2O
Chemistry
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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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![**Problem Statement:**
Calculate the percent yield when 28.16 g of \( \text{CO}_2 \) are formed from the reaction of 4.000 moles of \( \text{C}_8\text{H}_{18} \) with 8.000 moles of \( \text{O}_2 \).
**Chemical Reaction:**
\[ 2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O} \]
**Explanation:**
This balanced chemical equation represents the combustion of octane (\( \text{C}_8\text{H}_{18} \)) in oxygen (\( \text{O}_2 \)) to produce carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)).
**Steps to Calculate Percent Yield:**
1. **Calculate theoretical yield of \( \text{CO}_2 \):**
- Use stoichiometry based on the balanced equation to find the theoretical amount of \( \text{CO}_2 \) that should be produced with 4.000 moles of \( \text{C}_8\text{H}_{18} \) and 8.000 moles of \( \text{O}_2 \).
2. **Convert grams of \( \text{CO}_2 \) to moles:**
- Use the molar mass of \( \text{CO}_2 \) to convert 28.16 g to moles.
3. **Calculate the percent yield:**
- Use the formula:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield (in moles)}}{\text{Theoretical Yield (in moles)}} \right) \times 100\%
\]
These steps will allow you to determine how efficiently carbon dioxide was produced in the reaction considering both reactants' moles and the given mass of the product.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9b2b64b-ae9b-405a-bed2-c76b10e09818%2F99ad399e-470c-44d4-a170-9a2116c2ef82%2Fwbq96_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Calculate the percent yield when 28.16 g of \( \text{CO}_2 \) are formed from the reaction of 4.000 moles of \( \text{C}_8\text{H}_{18} \) with 8.000 moles of \( \text{O}_2 \).
**Chemical Reaction:**
\[ 2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O} \]
**Explanation:**
This balanced chemical equation represents the combustion of octane (\( \text{C}_8\text{H}_{18} \)) in oxygen (\( \text{O}_2 \)) to produce carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)).
**Steps to Calculate Percent Yield:**
1. **Calculate theoretical yield of \( \text{CO}_2 \):**
- Use stoichiometry based on the balanced equation to find the theoretical amount of \( \text{CO}_2 \) that should be produced with 4.000 moles of \( \text{C}_8\text{H}_{18} \) and 8.000 moles of \( \text{O}_2 \).
2. **Convert grams of \( \text{CO}_2 \) to moles:**
- Use the molar mass of \( \text{CO}_2 \) to convert 28.16 g to moles.
3. **Calculate the percent yield:**
- Use the formula:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield (in moles)}}{\text{Theoretical Yield (in moles)}} \right) \times 100\%
\]
These steps will allow you to determine how efficiently carbon dioxide was produced in the reaction considering both reactants' moles and the given mass of the product.
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