I have a reactor which is operating at steady state. The balanced reaction is shown below: 1 A + 5 B => 2 M + 1 N The feed to the reactor contains only A and B, and B is fed in excess by 14% of what is required to react all of the A. The reactor has a single outlet stream, and all unused reactants and all products leave the reactor in the same stream. All of the A fed to the reactor is consumed by the reaction. Calculate the mole fraction of M in the outlet from the reactor. Place your answer in the blank below, rounded to two decimal places.
I have a reactor which is operating at steady state. The balanced reaction is shown below: 1 A + 5 B => 2 M + 1 N The feed to the reactor contains only A and B, and B is fed in excess by 14% of what is required to react all of the A. The reactor has a single outlet stream, and all unused reactants and all products leave the reactor in the same stream. All of the A fed to the reactor is consumed by the reaction. Calculate the mole fraction of M in the outlet from the reactor. Place your answer in the blank below, rounded to two decimal places.
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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![### Reactor Operating at Steady State - Example Problem
**Balanced Reaction Equation:**
\[1A + 5B \rightarrow 2M + 1N\]
**Problem Description:**
A reactor is operating at a steady state. The feed to the reactor contains only reactants A and B, with B being fed in excess by 14% of what is required to react with all of A. The reactor has a single outlet stream, and all unused reactants and all products leave the reactor in the same stream.
All of the A fed to the reactor is consumed by the reaction.
**Objective:**
Calculate the mole fraction of M in the outlet from the reactor.
**Steps to Solve:**
1. Determine the moles of B fed to the reactor compared to A.
2. Use stoichiometry to find the amount of M produced.
3. Calculate the total moles in the outlet stream.
4. Calculate the mole fraction of M.
**Hint:**
- Remember, B is in excess by 14%, so for every mole of A, the moles of B are given by \(5 \times 1.14 = 5.70\) moles of B.
### Calculation Method:
1. Let's assume 1 mole of A is fed.
2. Moles of B fed = \(1 \times 5.7 = 5.7\) moles.
3. From the reaction, 1 mole of A produces 2 moles of M.
4. Since all A is consumed, moles of M produced = 2 moles.
5. Moles of N produced = 1 mole.
6. Total moles in the outlet stream:
\[ \text{Total moles} = \text{moles of M} + \text{moles of N} + \text{excess B} \]
Given data:
- Moles of M = 2
- Moles of N = 1
- Excess B = \( 5.7 - 5 \) moles = 0.7 moles
Thus,
Total moles = \(2 + 1 + 0.7 = 3.7\) moles
Mole fraction of M in the outlet stream:
\[ \text{Mole fraction of M} = \frac{\text{moles of M}}{\text{total moles}} = \frac{2}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42d7a143-2adc-420d-86c3-225c6cc98b5a%2F3ecc8542-54b1-4e20-aa72-0588cc0fb7b7%2F7inv23i_processed.png&w=3840&q=75)
Transcribed Image Text:### Reactor Operating at Steady State - Example Problem
**Balanced Reaction Equation:**
\[1A + 5B \rightarrow 2M + 1N\]
**Problem Description:**
A reactor is operating at a steady state. The feed to the reactor contains only reactants A and B, with B being fed in excess by 14% of what is required to react with all of A. The reactor has a single outlet stream, and all unused reactants and all products leave the reactor in the same stream.
All of the A fed to the reactor is consumed by the reaction.
**Objective:**
Calculate the mole fraction of M in the outlet from the reactor.
**Steps to Solve:**
1. Determine the moles of B fed to the reactor compared to A.
2. Use stoichiometry to find the amount of M produced.
3. Calculate the total moles in the outlet stream.
4. Calculate the mole fraction of M.
**Hint:**
- Remember, B is in excess by 14%, so for every mole of A, the moles of B are given by \(5 \times 1.14 = 5.70\) moles of B.
### Calculation Method:
1. Let's assume 1 mole of A is fed.
2. Moles of B fed = \(1 \times 5.7 = 5.7\) moles.
3. From the reaction, 1 mole of A produces 2 moles of M.
4. Since all A is consumed, moles of M produced = 2 moles.
5. Moles of N produced = 1 mole.
6. Total moles in the outlet stream:
\[ \text{Total moles} = \text{moles of M} + \text{moles of N} + \text{excess B} \]
Given data:
- Moles of M = 2
- Moles of N = 1
- Excess B = \( 5.7 - 5 \) moles = 0.7 moles
Thus,
Total moles = \(2 + 1 + 0.7 = 3.7\) moles
Mole fraction of M in the outlet stream:
\[ \text{Mole fraction of M} = \frac{\text{moles of M}}{\text{total moles}} = \frac{2}{
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