Interpretation Introduction To determine: The amount of CO in the flask. Explanation Given: Moles of H 2 are 0 .13 , moles of CO are 0 .13 , moles of H 2 O are 0 .43 . Explanation: The chemical reaction is as follows: H 2 O ( g ) + C ( s ) → H 2 ( g ) + CO ( g ) At equilibrium, Moles of H 2 are 0 .13 . Moles of CO are 0 .13 . Moles of H 2 O are 0 .43 . Calculate equilibrium constant ( K C ) as follows: K C = [ H 2 ] [ CO ] [ H 2 O ] (1) Substitute 0 .13 mol for H 2 , 0 .13 mol for CO , and 0 .43 mol for H 2 O in the above equation as follows: K C = [ 0 .13 mol ] [ 0 .13 mol ] [ 0 .43 mol ] = 0.0393 After reaction with O 2 , H 2 is completely reacted. Initial moles and final moles of chemical reaction will be as follows: H 2 O ( g ) + C ( s ) → H 2 ( g ) + CO ( g ) 0.43 mol 0 0 .13 mol − x x + x 0

Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package

1st Edition

ISBN: 9780321955517

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 14, Problem 76E

Interpretation Introduction

To determine: The amount of CO in the flask.

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Chapter 14, Problem 75E

Chapter 14, Problem 77E

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Chapter 14 Solutions

Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package

Ch. 14 - Prob. 1SAQ Ch. 14 - Q2. The equilibrium constant for the reaction...Ch. 14 - Q3. Use the data shown here to find the...Ch. 14 - Prob. 4SAQ Ch. 14 - Prob. 5SAQ Ch. 14 - Q6. For the reaction 2 A(g) B(g), the equilibrium...Ch. 14 - Q7. Consider the reaction between iodine gas and...Ch. 14 - Prob. 8SAQ Ch. 14 - Prob. 9SAQ Ch. 14 - Prob. 10SAQ

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