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Example 4.8. The combustion of n-heptane is C₂H₁6 + 110₂, 7CO₂ + 8H₂O Ten (10) kg of n-heptane is reacted with an excess amount of O₂, and 14.4 kg of CO, is formed. Calculate the conversion percentage of n-heptane. Since it is stated that O, is in excess, n-heptane is, therefore, a limiting reactant. The # of moles of 10 kg of C,H₁, fed and 14.4 kg of CO, generated can be computed as follows

Example 4.8. The combustion of n-heptane is C₂H₁6 + 110₂, 7CO₂ + 8H₂O Ten (10) kg of n-heptane is reacted with an excess amount of O₂, and 14.4 kg of CO, is formed. Calculate the conversion percentage of n-heptane. Since it is stated that O, is in excess, n-heptane is, therefore, a limiting reactant. The # of moles of 10 kg of C,H₁, fed and 14.4 kg of CO, generated can be computed as follows

Introduction to Chemical Engineering Thermodynamics
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 4.8. The combustion of n-heptane is C₂H16 + 110₂ 7CO₂ + 8H₂O Ten (10) kg of n-heptane is reacted with an excess amount of O₂, and 14.4 kg of CO₂ is formed. Calculate the conversion percentage of n-heptane. Since it is stated that O₂ is in excess, n-heptane is, therefore, a limiting reactant. The # of moles of 10 kg of C₂H16 fed and 14.4 kg of CO₂ generated can be computed as follows
Transcribed Image Text:Example 4.8. The combustion of n-heptane is C₂H16 + 110₂ 7CO₂ + 8H₂O Ten (10) kg of n-heptane is reacted with an excess amount of O₂, and 14.4 kg of CO₂ is formed. Calculate the conversion percentage of n-heptane. Since it is stated that O₂ is in excess, n-heptane is, therefore, a limiting reactant. The # of moles of 10 kg of C₂H16 fed and 14.4 kg of CO₂ generated can be computed as follows
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