Volume acetic acid: 20.0 mL Volume Isopentyl alcohol: 15.0 mL Mass of Isopentyl Acetate collected: 9.775 grams Density of acetic acid: 1.049 g/mL Density of Isopentyl alcohol: 0.813 g/mL
Volume acetic acid: 20.0 mL Volume Isopentyl alcohol: 15.0 mL Mass of Isopentyl Acetate collected: 9.775 grams Density of acetic acid: 1.049 g/mL Density of Isopentyl alcohol: 0.813 g/mL
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
Related questions
Question
please help find the Percent yeild and limiting regent given the collected data
Volume acetic acid: 20.0 mL
Volume Isopentyl alcohol: 15.0 mL
Mass of Isopentyl Acetate collected: 9.775 grams
Density of acetic acid: 1.049 g/mL
Density of Isopentyl alcohol: 0.813 g/mL
![**Percent Yield of Isoamyl Acetate Based on the Limiting Reagent**
**Topic: Chemistry | Subtopic: Stoichiometry and Limiting Reagents**
---
**Objective:**
Calculate the percent yield of isoamyl acetate based on the limiting reagent.
**Introduction:**
In chemical reactions, the limiting reagent dictates how much of the product can be produced. Here, we explore a reaction to determine the percent yield of isoamyl acetate (product B) using stoichiometric calculations.
**Reaction:**
\[ \text{Reaction: A} \rightarrow \text{B} \]
**Given Data:**
- Mass of reactant A: 5.00 g
- Molar mass of A: 136 g/mol
- Molar ratio of A to B: 1:1
- Molar mass of B: 181 g/mol
**Calculation:**
To find the theoretical yield of B:
1. Convert the mass of A to moles.
2. Use the molar ratio to convert moles of A to moles of B.
3. Convert moles of B to grams.
Mathematical setup:
\[ \text{5.00 g A} \times \left( \frac{1 \text{ mol A}}{136 \text{ g A}} \right) \times \left( \frac{1 \text{ mol B}}{1 \text{ mol A}} \right) \times \left( \frac{181 \text{ g B}}{1 \text{ mol B}} \right) = 6.65 \text{ g B} \]
This can be expressed as:
**Step-by-Step Calculation:**
1. **Convert grams of A to moles of A:**
\[
5.00 \text{ g A} \times \left( \frac{1 \text{ mol A}}{136 \text{ g A}} \right) = 0.03676 \text{ mol A}
\]
2. **Convert moles of A to moles of B** based on the given 1:1 molar ratio:
\[
0.03676 \text{ mol A} \times \left( \frac{1 \text{ mol B}}{1 \text{ mol A}} \right) = 0.03676 \text{ mol B}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6746fa6f-1856-443b-addc-ecae5d9da9b9%2Ff3745396-9b04-4944-bafa-d9e578c80f10%2F3chjbw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Percent Yield of Isoamyl Acetate Based on the Limiting Reagent**
**Topic: Chemistry | Subtopic: Stoichiometry and Limiting Reagents**
---
**Objective:**
Calculate the percent yield of isoamyl acetate based on the limiting reagent.
**Introduction:**
In chemical reactions, the limiting reagent dictates how much of the product can be produced. Here, we explore a reaction to determine the percent yield of isoamyl acetate (product B) using stoichiometric calculations.
**Reaction:**
\[ \text{Reaction: A} \rightarrow \text{B} \]
**Given Data:**
- Mass of reactant A: 5.00 g
- Molar mass of A: 136 g/mol
- Molar ratio of A to B: 1:1
- Molar mass of B: 181 g/mol
**Calculation:**
To find the theoretical yield of B:
1. Convert the mass of A to moles.
2. Use the molar ratio to convert moles of A to moles of B.
3. Convert moles of B to grams.
Mathematical setup:
\[ \text{5.00 g A} \times \left( \frac{1 \text{ mol A}}{136 \text{ g A}} \right) \times \left( \frac{1 \text{ mol B}}{1 \text{ mol A}} \right) \times \left( \frac{181 \text{ g B}}{1 \text{ mol B}} \right) = 6.65 \text{ g B} \]
This can be expressed as:
**Step-by-Step Calculation:**
1. **Convert grams of A to moles of A:**
\[
5.00 \text{ g A} \times \left( \frac{1 \text{ mol A}}{136 \text{ g A}} \right) = 0.03676 \text{ mol A}
\]
2. **Convert moles of A to moles of B** based on the given 1:1 molar ratio:
\[
0.03676 \text{ mol A} \times \left( \frac{1 \text{ mol B}}{1 \text{ mol A}} \right) = 0.03676 \text{ mol B}
![**Determining the Limiting Reagent:**
When conducting a chemical reaction, identifying the limiting reagent is crucial as it determines the maximum amount of product that can be formed. In this example, we will ignore sulfuric acid since it acts as a catalyst and does not affect the overall stoichiometry of the reaction.
**Example Reaction:**
For the general reaction:
\[ \text{A} + \text{B} \rightarrow \text{C} \]
**Calculating the Moles of Reactant A:**
Assume we have a reactant \( \text{A} \) with a mass of 20.00 grams. To convert this mass into moles, use the molar mass (MW) of \( \text{A} \). The number of moles of \( \text{A} \) is calculated as follows:
\[ 20.00 \text{ g A} \times \left( \frac{1 \text{ mol A}}{\text{MW A}} \right) \times 1 = \text{mole ratio A} \]
This can be represented more generally as:
\[ 20.00 \text{ g A} \times \left( \frac{1 \text{ mol A}}{\text{MW A}} \right) \times \left( \frac{1}{1 \text{ mol A}} \right) = \text{mole ratio A} \]
Here’s the breakdown:
1. **Mass of A:** \( 20.00 \) grams
2. **Conversion Factor:** \(\left( \frac{1 \text{ mol A}}{\text{MW A}} \right)\) (The reciprocal of the molar mass of \( \text{A} \)),
3. **Stoichiometric Coefficient:** \(\left( \frac{1}{1 \text{ mol A}} \right)\) (assuming a 1:1 mole ratio from the balanced equation).
By substituting the molar mass of A (MW A) into the equation, you can calculate the mole ratio of A, which will aid in determining if \( \text{A} \) is the limiting reagent in this reaction. This method can be applied similarly to reactant \( \text{B} \) to compare which reactant will limit the formation of product \( \text{C} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6746fa6f-1856-443b-addc-ecae5d9da9b9%2Ff3745396-9b04-4944-bafa-d9e578c80f10%2Fkjkwww_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Determining the Limiting Reagent:**
When conducting a chemical reaction, identifying the limiting reagent is crucial as it determines the maximum amount of product that can be formed. In this example, we will ignore sulfuric acid since it acts as a catalyst and does not affect the overall stoichiometry of the reaction.
**Example Reaction:**
For the general reaction:
\[ \text{A} + \text{B} \rightarrow \text{C} \]
**Calculating the Moles of Reactant A:**
Assume we have a reactant \( \text{A} \) with a mass of 20.00 grams. To convert this mass into moles, use the molar mass (MW) of \( \text{A} \). The number of moles of \( \text{A} \) is calculated as follows:
\[ 20.00 \text{ g A} \times \left( \frac{1 \text{ mol A}}{\text{MW A}} \right) \times 1 = \text{mole ratio A} \]
This can be represented more generally as:
\[ 20.00 \text{ g A} \times \left( \frac{1 \text{ mol A}}{\text{MW A}} \right) \times \left( \frac{1}{1 \text{ mol A}} \right) = \text{mole ratio A} \]
Here’s the breakdown:
1. **Mass of A:** \( 20.00 \) grams
2. **Conversion Factor:** \(\left( \frac{1 \text{ mol A}}{\text{MW A}} \right)\) (The reciprocal of the molar mass of \( \text{A} \)),
3. **Stoichiometric Coefficient:** \(\left( \frac{1}{1 \text{ mol A}} \right)\) (assuming a 1:1 mole ratio from the balanced equation).
By substituting the molar mass of A (MW A) into the equation, you can calculate the mole ratio of A, which will aid in determining if \( \text{A} \) is the limiting reagent in this reaction. This method can be applied similarly to reactant \( \text{B} \) to compare which reactant will limit the formation of product \( \text{C} \).
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 8 steps with 10 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemical-engineering and related others by exploring similar questions and additional content below.Recommended textbooks for you
Introduction to Chemical Engineering Thermodynami…
Chemical Engineering
ISBN:
9781259696527
Author:
J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:
McGraw-Hill Education
Elementary Principles of Chemical Processes, Bind…
Chemical Engineering
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
Elements of Chemical Reaction Engineering (5th Ed…
Chemical Engineering
ISBN:
9780133887518
Author:
H. Scott Fogler
Publisher:
Prentice Hall
Introduction to Chemical Engineering Thermodynami…
Chemical Engineering
ISBN:
9781259696527
Author:
J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:
McGraw-Hill Education
Elementary Principles of Chemical Processes, Bind…
Chemical Engineering
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
Elements of Chemical Reaction Engineering (5th Ed…
Chemical Engineering
ISBN:
9780133887518
Author:
H. Scott Fogler
Publisher:
Prentice Hall
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:
9781285061238
Author:
Lokensgard, Erik
Publisher:
Delmar Cengage Learning
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:
9780072848236
Author:
Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:
McGraw-Hill Companies, The