I have a reaction occurring at steady state, but the feed is an unknown composition. The reaction occurs according to the balanced equation below: 1 A + 2B => 1 M + 3 N In the outlet stream, the amount of B is equal to the amount of N. Also, the amount of M is 150% of the amount of A in the outlet stream. Find the mole fraction of A in the feed stream, and the fractional conversion of A in the reaction.
I have a reaction occurring at steady state, but the feed is an unknown composition. The reaction occurs according to the balanced equation below: 1 A + 2B => 1 M + 3 N In the outlet stream, the amount of B is equal to the amount of N. Also, the amount of M is 150% of the amount of A in the outlet stream. Find the mole fraction of A in the feed stream, and the fractional conversion of A in the reaction.
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
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![### Example Problem on Reaction Stoichiometry and Mole Fraction
**Problem Statement:**
"I have a reaction occurring at steady state, but the feed is an unknown composition. The reaction occurs according to the balanced equation below:
\[ 1A + 2B \rightarrow 1M + 3N \]
In the outlet stream, the amount of B is equal to the amount of N. Also, the amount of M is 150% of the amount of A in the outlet stream.
**Questions:**
1. Find the mole fraction of A in the feed stream.
2. Determine the fractional conversion of A in the reaction."
### Breakdown and Explanation:
Given data from the problem:
- **Balanced Reaction Equation:**
\[ 1A + 2B \rightarrow 1M + 3N \]
- In the outlet stream:
- The amount of \( B \) is equal to the amount of \( N \).
- The amount of \( M \) is 150% of the amount of \( A \).
To solve these types of problems, we must use stoichiometry along with the given information about the outlet stream composition to find the unknowns.
#### Step-by-Step Solution:
1. **Establish relationships based on the stoichiometry:**
- Outlet stream composition relationships:
\[ n_B = n_N \]
\[ n_M = 1.5 n_A \]
2. **Using the balanced equation, set up mole balances:**
- Let's denote \( n_A \) as the moles of A, \( n_B \) as the moles of B, \( n_M \) as the moles of M, and \( n_N \) as the moles of N in the outlet stream.
3. **Using the given relationships:**
- Since the amount of B is equal to the amount of N:
\[ n_B = n_N \]
- Since the amount of M is 150% of the amount of A:
\[ n_M = 1.5 n_A \]
4. **Find the mole fraction of A in the feed stream:**
- Let the feed stream contain a total of \( F \) moles.
- Mole fraction of A in the feed stream:
\[ x_A = \frac{n_A}{F} \]
To find \( F \), we](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42d7a143-2adc-420d-86c3-225c6cc98b5a%2F44d3cf5f-55bb-4a3c-8561-b9f014b44cec%2Fh97pdlb_processed.png&w=3840&q=75)
Transcribed Image Text:### Example Problem on Reaction Stoichiometry and Mole Fraction
**Problem Statement:**
"I have a reaction occurring at steady state, but the feed is an unknown composition. The reaction occurs according to the balanced equation below:
\[ 1A + 2B \rightarrow 1M + 3N \]
In the outlet stream, the amount of B is equal to the amount of N. Also, the amount of M is 150% of the amount of A in the outlet stream.
**Questions:**
1. Find the mole fraction of A in the feed stream.
2. Determine the fractional conversion of A in the reaction."
### Breakdown and Explanation:
Given data from the problem:
- **Balanced Reaction Equation:**
\[ 1A + 2B \rightarrow 1M + 3N \]
- In the outlet stream:
- The amount of \( B \) is equal to the amount of \( N \).
- The amount of \( M \) is 150% of the amount of \( A \).
To solve these types of problems, we must use stoichiometry along with the given information about the outlet stream composition to find the unknowns.
#### Step-by-Step Solution:
1. **Establish relationships based on the stoichiometry:**
- Outlet stream composition relationships:
\[ n_B = n_N \]
\[ n_M = 1.5 n_A \]
2. **Using the balanced equation, set up mole balances:**
- Let's denote \( n_A \) as the moles of A, \( n_B \) as the moles of B, \( n_M \) as the moles of M, and \( n_N \) as the moles of N in the outlet stream.
3. **Using the given relationships:**
- Since the amount of B is equal to the amount of N:
\[ n_B = n_N \]
- Since the amount of M is 150% of the amount of A:
\[ n_M = 1.5 n_A \]
4. **Find the mole fraction of A in the feed stream:**
- Let the feed stream contain a total of \( F \) moles.
- Mole fraction of A in the feed stream:
\[ x_A = \frac{n_A}{F} \]
To find \( F \), we
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