Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card
Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card
11th Edition
ISBN: 9781337128391
Author: Darrell Ebbing, Steven D. Gammon
Publisher: Cengage Learning
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Chapter 23, Problem 23.31QP

Complete and balance the following equations. Note any catalyst used.

  1. a C 2 H 4 + O 2
  2. b CH 2 = CH 2 + MnO 4 + H 2 O
  3. c CH 2 = CH 2 + Br 2
  4. d Chapter 23, Problem 23.31QP, Complete and balance the following equations. Note any catalyst used. a C2H4+O2 b CH2=CH2+MnO4+H2O c , example  1
  5. e Chapter 23, Problem 23.31QP, Complete and balance the following equations. Note any catalyst used. a C2H4+O2 b CH2=CH2+MnO4+H2O c , example  2

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

For the given reactions, the equation has to be completed and balanced.

Concept Introduction:

Combustion reaction is the one in which, the substance is burnt in presence of oxygen.  If any hydrocarbon is burnt in presence of oxygen, the products obtained is carbon dioxide and water only.

Answer to Problem 23.31QP

The balanced equation for the given reaction can be written as,

C2H4+ 3O2  2CO2+ 2H2O

Explanation of Solution

The given reaction in the problem statement is a combustion reaction and hence, the product obtained will be carbon dioxide and water only.  Hence, the raw equation can be written as,

C2H4+ O2  CO2+ H2O

As the total number of atoms present in both reactants and products must be equal, balancing has to be done.  To balance the carbon atom in both sides, the product side has to be multiplied by 2.  This leads, balancing of both carbon and hydrogen atoms on both sides.

C2H4+ O2  2CO2+ 2H2O

Oxygen count was six in product side.  But in the reactant side it is two.  Hence, multiplying by three on reactant side balances the oxygen atom also on both sides.  Therefore, the balanced chemical equation for the given reaction can be given as,

C2H4+ 3O2  2CO2+ 2H2O

Conclusion

The given incomplete combustion reaction is completed and the balanced chemical equation also written.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

For the given reactions, the equation has to be completed and balanced.

Concept Introduction:

Oxidation is loss of electron by a species.  This also can be said as addition of oxygen.  Reduction is a process in which electrons are gained by a species.

Answer to Problem 23.31QP

The balanced equation for the given reaction can be written as,

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  1

Explanation of Solution

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  2

Reduction:  MnO4-  MnO2        Balance O:  MnO4-+ 2H2 MnO2+ 4OH-        Balance e-:  MnO4-+ 2H2O + 3e-  MnO2+ 4OH-

Add half-reactions:

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  3

Therefore, the complete equation for the given reaction can be given as,

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  4

Conclusion

The balanced equation for the given reaction was written.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

For the given reactions, the equation has to be completed and balanced.

Concept Introduction:

Bromination of alkene happens without any catalyst in carbon tetrachloride.  The bromine atom adds across the double bond reducing the double bond to a single bond.

Answer to Problem 23.31QP

The balanced equation for the given reaction can be written as,

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  5

Explanation of Solution

Bromination across the double bond.

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  6

Conclusion

The given bromination reaction was completed and the product was written with balanced equation.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

For the given reactions, the equation has to be completed and balanced.

Concept Introduction:

Bromination of aromatic compound happens in presence of catalyst.  Usually this does not occur in absence of catalyst.  The position of bromination in the aromatic ring depends upon the group already substituted in the ring.

Answer to Problem 23.31QP

The balanced equation for the given reaction can be written as,

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  7

Explanation of Solution

Bromination of toluene takes place in presence of catalyst with bromine.  As methyl is a para directing group, the bromine atom is substituted in the para position.  By considering this, the complete reaction can be given as,

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  8

Conclusion

The given bromination reaction was completed and the product was written with balanced equation.

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

For the given reactions, the equation has to be completed and balanced.

Concept Introduction:

Nitration of aromatic compound happens when concentrated nitric acid and sulphuric acid is added to it.  The position of nitro group substitution depends upon the position of the already substituted group present in it.

Answer to Problem 23.31QP

The balanced equation for the given reaction can be written as,

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  9

Explanation of Solution

Nitration of toluene takes place in presence of nitric acid and sulphuric acid.  As methyl is a para directing group, the nitro group is substituted in the para position.  By considering this, the complete reaction can be given as,

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card, Chapter 23, Problem 23.31QP , additional homework tip  10

Conclusion

The given reaction was completed and the product was written with balanced equation.

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Chapter 23 Solutions

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card

Ch. 23.5 - Give the IUPAC name for each of the following...Ch. 23.5 - Prob. 23.10ECh. 23.5 - Prob. 23.3CCCh. 23.6 - Prob. 23.11ECh. 23.6 - Prob. 23.12ECh. 23.6 - Prob. 23.13ECh. 23 - Give the molecular formula of an alkane with 25...Ch. 23 - Prob. 23.2QPCh. 23 - Prob. 23.3QPCh. 23 - Prob. 23.4QPCh. 23 - Prob. 23.5QPCh. 23 - Prob. 23.6QPCh. 23 - Prob. 23.7QPCh. 23 - Prob. 23.8QPCh. 23 - What would you expect to be the major product when...Ch. 23 - Prob. 23.10QPCh. 23 - Prob. 23.11QPCh. 23 - Prob. 23.12QPCh. 23 - Prob. 23.13QPCh. 23 - Prob. 23.14QPCh. 23 - Prob. 23.15QPCh. 23 - Prob. 23.16QPCh. 23 - Prob. 23.17QPCh. 23 - What is the correct IUPAC name for the following...Ch. 23 - Prob. 23.19QPCh. 23 - Prob. 23.20QPCh. 23 - Explain why you wouldnt expect to find a compound...Ch. 23 - Catalytic cracking is an industrial process used...Ch. 23 - Prob. 23.23QPCh. 23 - In the models shown here, C atoms are black and H...Ch. 23 - Prob. 23.25QPCh. 23 - Prob. 23.26QPCh. 23 - Prob. 23.27QPCh. 23 - Prob. 23.28QPCh. 23 - Prob. 23.29QPCh. 23 - Prob. 23.30QPCh. 23 - Complete and balance the following equations. Note...Ch. 23 - Prob. 23.32QPCh. 23 - Prob. 23.33QPCh. 23 - Prob. 23.34QPCh. 23 - Prob. 23.35QPCh. 23 - Complete the following equation, giving only the...Ch. 23 - Prob. 23.37QPCh. 23 - What is the IUPAC name of each of the following...Ch. 23 - Prob. 23.39QPCh. 23 - Write the condensed structural formula for each of...Ch. 23 - Give the IUPAC name of each of the following. a...Ch. 23 - For each of the following, write the IUPAC name. a...Ch. 23 - Prob. 23.43QPCh. 23 - Prob. 23.44QPCh. 23 - Prob. 23.45QPCh. 23 - Prob. 23.46QPCh. 23 - Give the IUPAC name of each of the following...Ch. 23 - Prob. 23.48QPCh. 23 - Prob. 23.49QPCh. 23 - Prob. 23.50QPCh. 23 - Prob. 23.51QPCh. 23 - Circle and name the functional group in each...Ch. 23 - Prob. 23.53QPCh. 23 - Prob. 23.54QPCh. 23 - Prob. 23.55QPCh. 23 - Prob. 23.56QPCh. 23 - What is the common name of each of the following...Ch. 23 - Prob. 23.58QPCh. 23 - Prob. 23.59QPCh. 23 - Prob. 23.60QPCh. 23 - Prob. 23.61QPCh. 23 - Prob. 23.62QPCh. 23 - Give the IUPAC name of each of the following...Ch. 23 - Prob. 23.64QPCh. 23 - Prob. 23.65QPCh. 23 - Prob. 23.66QPCh. 23 - Prob. 23.67QPCh. 23 - Prob. 23.68QPCh. 23 - Prob. 23.69QPCh. 23 - Prob. 23.70QPCh. 23 - Prob. 23.71QPCh. 23 - A compound with a fragrant odor reacts with dilute...Ch. 23 - Prob. 23.73QPCh. 23 - Prob. 23.74QPCh. 23 - Prob. 23.75QPCh. 23 - Prob. 23.76QPCh. 23 - Prob. 23.77QPCh. 23 - Prob. 23.78QPCh. 23 - Prob. 23.79QPCh. 23 - Prob. 23.80QPCh. 23 - Prob. 23.81QPCh. 23 - Prob. 23.82QPCh. 23 - Prob. 23.83QPCh. 23 - Prob. 23.84QPCh. 23 - Prob. 23.85QPCh. 23 - Prob. 23.86QPCh. 23 - Prob. 23.87QPCh. 23 - Prob. 23.88QP
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