Homework Help is Here – Start Your Trial Now!
Science
Chemistry
At the end of this, how did we go from a 5 to 1 ratio moles to 2CO2 and 3H2O?

At the end of this, how did we go from a 5 to 1 ratio moles to 2CO2 and 3H2O?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
See similar textbooks
icon
Related questions
Question
100%

At the end of this, how did we go from a 5 to 1 ratio moles to 2CO2 and 3H2O?

Number of moles of hydrocarbon is calculated as follows: PV = nRT Rearrange above equation for n. PV = nRT n = RT Value of Vis V. Value of R is 0.082057 L atm K-¹.mol-¹. Value of Tis 298 K. Value of P is 0.959 atm. Substitute the values in above equation for moles of hydrocarbon. n = RT (0.959 atm) (V) (0.082057 L-atm-K-¹-mol-¹) (298 K) = 0.0392 V mol Hence, number of moles of hydrocarbon is 0.0392 V mol. Value of Vis 4V. Value of R is 0.082057 L atm K-¹ mol-¹. Value of Tis 375 K. Value of P is 1.51 atm. Substitute the values in above equation for moles of mixture of gases. n = RT (1.51 atm)(4V) (0.082057 L-atm-K-¹-mol-¹)(375 K) = 0.1962 V mol Hence, number of moles of carbon dioxide and water vapor mixture is 0.1962 V mol. Ratio of number of moles of hydrocarbons to mixture of gases is as follows: number of moles of hydrocarbon Ratio = number of moles of mixture Value of number of moles of hydrocarbon is 0.0392 V mol. Value of number of moles of mixture is 0.1962 V mol. Substitute values in above expression. Ratio = number of moles of mixture number of moles of hydrocarbon = Conclusion 0.1962 V mol 0.0392 V mol Hence, total number of moles of mixture of carbon dioxide and water vapor is 5 and hydrocarbon is 1. Therefore, combustion reaction of hydrocarbon so formed is as follows: Hydrocarbon + O2 → 2CO2 + 3H₂O From above reaction it is evident that number of carbon and hydrogen in molecular formula of hydrocarbon is 2 and 6. Hence, molecular formula of hydrocarbon is C₂H6. Molecular formula of hydrocarbon is C₂H6.
Transcribed Image Text:Number of moles of hydrocarbon is calculated as follows: PV = nRT Rearrange above equation for n. PV = nRT n = RT Value of Vis V. Value of R is 0.082057 L atm K-¹.mol-¹. Value of Tis 298 K. Value of P is 0.959 atm. Substitute the values in above equation for moles of hydrocarbon. n = RT (0.959 atm) (V) (0.082057 L-atm-K-¹-mol-¹) (298 K) = 0.0392 V mol Hence, number of moles of hydrocarbon is 0.0392 V mol. Value of Vis 4V. Value of R is 0.082057 L atm K-¹ mol-¹. Value of Tis 375 K. Value of P is 1.51 atm. Substitute the values in above equation for moles of mixture of gases. n = RT (1.51 atm)(4V) (0.082057 L-atm-K-¹-mol-¹)(375 K) = 0.1962 V mol Hence, number of moles of carbon dioxide and water vapor mixture is 0.1962 V mol. Ratio of number of moles of hydrocarbons to mixture of gases is as follows: number of moles of hydrocarbon Ratio = number of moles of mixture Value of number of moles of hydrocarbon is 0.0392 V mol. Value of number of moles of mixture is 0.1962 V mol. Substitute values in above expression. Ratio = number of moles of mixture number of moles of hydrocarbon = Conclusion 0.1962 V mol 0.0392 V mol Hence, total number of moles of mixture of carbon dioxide and water vapor is 5 and hydrocarbon is 1. Therefore, combustion reaction of hydrocarbon so formed is as follows: Hydrocarbon + O2 → 2CO2 + 3H₂O From above reaction it is evident that number of carbon and hydrogen in molecular formula of hydrocarbon is 2 and 6. Hence, molecular formula of hydrocarbon is C₂H6. Molecular formula of hydrocarbon is C₂H6.
Expert Solution
Step 1

Chemistry homework question answer, step 1, image 1

trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Mole Concept
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
SEE MORE TEXTBOOKS