in determine the molar amounts atomic species balancer and • molecular specier balancen
in determine the molar amounts atomic species balancer and • molecular specier balancen
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
Demonstrate the molecular species balance on this reaction
![**Propylene Ammoxidation Process**
The chemical reaction for the propylene ammoxidation process is given by:
\[ \text{C}_3\text{H}_6 + \text{NH}_3 + 3/2 \text{O}_2 \rightarrow \text{C}_3\text{H}_3\text{N} + 3\text{H}_2\text{O} \]
**Problem Statement:**
The feed to the propylene ammoxidation process contains:
- 10.0% mole propylene (C₃H₆)
- 12.0% ammonia (NH₃)
- 78.0% air
A fractional conversion of 30.0% of the limiting reactant is achieved. Given 100 mol of feed to the reactor, determine:
1. The limiting reactant.
2. The molar amount of all gas constituents for a 30% conversion of the limiting reactant.
3. Determine the molar amounts using atomic species balances and molecular species balances.
**Reactor Setup:**
A box diagram represents the reactor with input and output streams.
**Feed:**
- Total: 100 mol
- Composition:
- 10.0 mol of C₃H₆
- 12.0 mol of NH₃
- 23.1 mol of O₂ (from air)
- 56.9 mol of N₂ (from air)
**Outputs:**
- \( n_{\text{C}_3\text{H}_6} \): C₃H₆
- \( n_{\text{NH}_3} \): NH₃
- \( n_{\text{O}_2} \): O₂
- \( n_{\text{N}_2} \): N₂
- \( n_{\text{C}_3\text{H}_3\text{N}} \): C₃H₃N
- \( n_{\text{H}_2\text{O}} \): H₂O
**Answers:**
1. \( n_{\text{C}_3\text{H}_6} = 7.0 \, \text{mol} \, \text{C}_3\text{H}_6 \)
2. \( n_{\text{NH}_3} = 9.9 \, \text{mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ba88ecd-654b-4f6b-9783-e93c33ca6a16%2Faab52e4b-802f-4959-ad18-028cadbe5105%2Ff60ob9y_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Propylene Ammoxidation Process**
The chemical reaction for the propylene ammoxidation process is given by:
\[ \text{C}_3\text{H}_6 + \text{NH}_3 + 3/2 \text{O}_2 \rightarrow \text{C}_3\text{H}_3\text{N} + 3\text{H}_2\text{O} \]
**Problem Statement:**
The feed to the propylene ammoxidation process contains:
- 10.0% mole propylene (C₃H₆)
- 12.0% ammonia (NH₃)
- 78.0% air
A fractional conversion of 30.0% of the limiting reactant is achieved. Given 100 mol of feed to the reactor, determine:
1. The limiting reactant.
2. The molar amount of all gas constituents for a 30% conversion of the limiting reactant.
3. Determine the molar amounts using atomic species balances and molecular species balances.
**Reactor Setup:**
A box diagram represents the reactor with input and output streams.
**Feed:**
- Total: 100 mol
- Composition:
- 10.0 mol of C₃H₆
- 12.0 mol of NH₃
- 23.1 mol of O₂ (from air)
- 56.9 mol of N₂ (from air)
**Outputs:**
- \( n_{\text{C}_3\text{H}_6} \): C₃H₆
- \( n_{\text{NH}_3} \): NH₃
- \( n_{\text{O}_2} \): O₂
- \( n_{\text{N}_2} \): N₂
- \( n_{\text{C}_3\text{H}_3\text{N}} \): C₃H₃N
- \( n_{\text{H}_2\text{O}} \): H₂O
**Answers:**
1. \( n_{\text{C}_3\text{H}_6} = 7.0 \, \text{mol} \, \text{C}_3\text{H}_6 \)
2. \( n_{\text{NH}_3} = 9.9 \, \text{mol
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