Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
Book Icon
Chapter 10, Problem 10.108QA
Interpretation Introduction

To find:

The partial pressure of CO

Expert Solution & Answer
Check Mark

Answer to Problem 10.108QA

Solution:

The partial pressure of CO is 0.320 atm

Explanation of Solution

1) Concept:

We are asked to find the partial pressure of CO gas in the reaction mixture. We need to find first the limiting reactant. Then, we need to find moles of gaseous products CO and CO2 that are formed during the reaction and moles of unreacted  O2. Combining the moles will give us the total moles of gaseous species. Using the total moles of gaseous products and moles of O, we can find mole fraction of CO.

According to Dalton’s law of partial pressure, the partial pressure of an individual is equal to the product of its mole fraction and total pressure.

i.e.,

Pi=χi×Ptotal

where, Pi= partial pressure of component  i, χi= mole fraction of component  i

Ptotal= total pressure.

Mole fraction is the ratio of moles of individual component to the total moles in the mixture.

i.e.

χA=mol of A(mol of A+mol of B)=mol of Atotal mol

2) Formula:

i) Pi=χi×Ptotal

ii) χA=mol of A(mol of A+mol of B)

3) Given:

i) 0.156 mol C

ii) 0.117 mol O2

iii) V=10.0 L

iv) T=500 K

v) Ptotal=0.640 atm

4) Calculation:

i) The balanced equation for the reaction is

4C + 3O2 → 2CO2 + 2 CO

Finding the limiting reactant as follows:

0.156 mol C ×3 mol O24 mol C=0.117 mol

As the two reactants are in stoichiometric amounts, both are completely consumed. So, after the reaction is complete, only gaseous products are present, which will govern the total pressure.

ii) From the stoichiometry of the reaction, one mole of C produces two moles of CO2, so converting moles of C into moles of CO2 as

0.156 mol C×1 mol CO2 2 mol C=0.0780 mol CO2

From the stoichiometry of the reaction, one mole of C produces two moles of CO , so converting moles of C into moles of CO  as

0.156 mol C×1 mol CO 2 mol C=0.0780 mol CO

Calculating the mole fraction of CO as

χCO=mol of COmol of CO+mol of CO2

χCO=0.0780 mol (0.0780mol+0.0780 mol )=0.500

Calculating the partial pressure of CO

PCO=χCOi×Ptotal

PCO=0.500×0.640 atm=0.320 atm

Conclusion:

Dalton’s law of partial pressure is used to find partial pressure of each of the gaseous compound in a given reaction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Previous
Chapter 10, Problem 10.107QA
Next
Chapter 10, Problem 10.109QA

Chapter 10 Solutions

Chemistry: An Atoms-Focused Approach

Ch. 10 - Prob. 10.1VPCh. 10 - Prob. 10.2VPCh. 10 - Prob. 10.3VPCh. 10 - Prob. 10.4VPCh. 10 - Prob. 10.6VPCh. 10 - Prob. 10.7VPCh. 10 - Prob. 10.8VPCh. 10 - Prob. 10.9VPCh. 10 - Prob. 10.10VPCh. 10 - Prob. 10.11VP
Ch. 10 - Prob. 10.13VPCh. 10 - Prob. 10.14VPCh. 10 - Prob. 10.15VPCh. 10 - Prob. 10.16VPCh. 10 - Prob. 10.17VPCh. 10 - Prob. 10.18VPCh. 10 - Prob. 10.19VPCh. 10 - Prob. 10.20VPCh. 10 - Prob. 10.21QACh. 10 - Prob. 10.22QACh. 10 - Prob. 10.23QACh. 10 - Prob. 10.24QACh. 10 - Prob. 10.25QACh. 10 - Prob. 10.26QACh. 10 - Prob. 10.27QACh. 10 - Prob. 10.28QACh. 10 - Prob. 10.29QACh. 10 - Prob. 10.30QACh. 10 - Prob. 10.31QACh. 10 - Prob. 10.32QACh. 10 - Prob. 10.33QACh. 10 - Prob. 10.34QACh. 10 - Prob. 10.35QACh. 10 - Prob. 10.36QACh. 10 - Prob. 10.37QACh. 10 - Prob. 10.38QACh. 10 - Prob. 10.39QACh. 10 - Prob. 10.40QACh. 10 - Prob. 10.41QACh. 10 - Prob. 10.42QACh. 10 - Prob. 10.43QACh. 10 - Prob. 10.44QACh. 10 - Prob. 10.45QACh. 10 - Prob. 10.46QACh. 10 - Prob. 10.47QACh. 10 - Prob. 10.48QACh. 10 - Prob. 10.49QACh. 10 - Prob. 10.50QACh. 10 - Prob. 10.51QACh. 10 - Prob. 10.52QACh. 10 - Prob. 10.53QACh. 10 - Prob. 10.54QACh. 10 - Prob. 10.55QACh. 10 - Prob. 10.56QACh. 10 - Prob. 10.57QACh. 10 - Prob. 10.58QACh. 10 - Prob. 10.59QACh. 10 - Prob. 10.60QACh. 10 - Prob. 10.61QACh. 10 - Prob. 10.62QACh. 10 - Prob. 10.63QACh. 10 - Prob. 10.64QACh. 10 - Prob. 10.65QACh. 10 - Prob. 10.66QACh. 10 - Prob. 10.67QACh. 10 - Prob. 10.68QACh. 10 - Prob. 10.69QACh. 10 - Prob. 10.70QACh. 10 - Prob. 10.71QACh. 10 - Prob. 10.72QACh. 10 - Prob. 10.73QACh. 10 - Prob. 10.74QACh. 10 - Prob. 10.75QACh. 10 - Prob. 10.76QACh. 10 - Prob. 10.77QACh. 10 - Prob. 10.78QACh. 10 - Prob. 10.79QACh. 10 - Prob. 10.80QACh. 10 - Prob. 10.81QACh. 10 - Prob. 10.82QACh. 10 - Prob. 10.83QACh. 10 - Prob. 10.84QACh. 10 - Prob. 10.85QACh. 10 - Prob. 10.86QACh. 10 - Prob. 10.87QACh. 10 - Prob. 10.88QACh. 10 - Prob. 10.89QACh. 10 - Prob. 10.90QACh. 10 - Prob. 10.91QACh. 10 - Prob. 10.92QACh. 10 - Prob. 10.93QACh. 10 - Prob. 10.94QACh. 10 - Prob. 10.95QACh. 10 - Prob. 10.96QACh. 10 - Prob. 10.97QACh. 10 - Prob. 10.98QACh. 10 - Prob. 10.99QACh. 10 - Prob. 10.100QACh. 10 - Prob. 10.101QACh. 10 - Prob. 10.102QACh. 10 - Prob. 10.103QACh. 10 - Prob. 10.104QACh. 10 - Prob. 10.105QACh. 10 - Prob. 10.106QACh. 10 - Prob. 10.107QACh. 10 - Prob. 10.108QACh. 10 - Prob. 10.109QACh. 10 - Prob. 10.110QACh. 10 - Prob. 10.111QACh. 10 - Prob. 10.112QACh. 10 - Prob. 10.113QACh. 10 - Prob. 10.114QACh. 10 - Prob. 10.115QACh. 10 - Prob. 10.116QACh. 10 - Prob. 10.117QACh. 10 - Prob. 10.118QACh. 10 - Prob. 10.119QACh. 10 - Prob. 10.120QACh. 10 - Prob. 10.121QACh. 10 - Prob. 10.122QACh. 10 - Prob. 10.123QACh. 10 - Prob. 10.124QACh. 10 - Prob. 10.125QACh. 10 - Prob. 10.126QACh. 10 - Prob. 10.127QACh. 10 - Prob. 10.128QACh. 10 - Prob. 10.129QACh. 10 - Prob. 10.130QACh. 10 - Prob. 10.131QACh. 10 - Prob. 10.132QACh. 10 - Prob. 10.133QACh. 10 - Prob. 10.134QACh. 10 - Prob. 10.135QACh. 10 - Prob. 10.136QACh. 10 - Prob. 10.137QACh. 10 - Prob. 10.138QACh. 10 - Prob. 10.139QACh. 10 - Prob. 10.140QACh. 10 - Prob. 10.141QACh. 10 - Prob. 10.142QACh. 10 - Prob. 10.143QACh. 10 - Prob. 10.144QACh. 10 - Prob. 10.145QACh. 10 - Prob. 10.146QACh. 10 - Prob. 10.147QACh. 10 - Prob. 10.148QACh. 10 - Prob. 10.149QACh. 10 - Prob. 10.150QA
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
SEE MORE TEXTBOOKS