INTRODUCTORY CHEMISTRY-STD.GDE.+SOL.MAN
INTRODUCTORY CHEMISTRY-STD.GDE.+SOL.MAN
8th Edition
ISBN: 9780134580289
Author: CORWIN
Publisher: PEARSON
Question
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Chapter 7, Problem 39E
Interpretation Introduction

(a)

Interpretation:

The balanced equation for the decomposition reaction of solid cobalt (III) carbonate is to be stated.

Concept introduction:

A chemical reaction is a process in which rearrangement of atoms or ions takes place between two reacting species. A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on both sides of the equation is equal.

Expert Solution
Check Mark

Answer to Problem 39E

The balanced equation for the decomposition reaction of solid cobalt (III) carbonate is shown below.

Co2(CO3)3(s)Co2O3(s)+3CO2(g)

Explanation of Solution

Solid cobalt (III) carbonate undergoes a decomposition reaction to produce solid cobalt (III) oxide, and carbon dioxide gas.

The unbalanced equation for the decomposition reaction of solid cobalt (III) carbonate is shown below.

Co2(CO3)3(s)Co2O3(s)+CO2(g)

The numbers of atoms of elements on the left side of the reaction are not equal to the right side of the reaction. Therefore, the reaction is unbalanced.

A coefficient of 3 is placed in front of CO2 to balance the chemical equation.

Therefore, the balanced equation for the decomposition reaction of solid cobalt (III) carbonate is shown below.

Co2(CO3)3(s)Co2O3(s)+3CO2(g)

The numbers of atoms of each element on the left side of the reaction are equal to that on the right side of the reaction. Therefore, the reaction is balanced.

Conclusion

The balanced equation for the decomposition reaction of solid cobalt (III) carbonate is shown below.

Co2(CO3)3(s)Co2O3(s)+3CO2(g)

Interpretation Introduction

(b)

Interpretation:

The balanced equation for the decomposition reaction of solid tin (IV) carbonate is to be stated.

Concept introduction:

A chemical reaction is a process in which rearrangement of atoms or ions takes place between two reacting species. A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on both sides of the equation is equal.

Expert Solution
Check Mark

Answer to Problem 39E

The balanced equation for the decomposition reaction of solid tin (IV) carbonate is shown below.

Sn(CO3)2(s)SnO2(s)+2CO2(g)

Explanation of Solution

Solid tin (IV) carbonate undergoes a decomposition reaction to produce tin (IV) oxide and carbon dioxide gas.

The unbalanced equation for the decomposition reaction of solid tin (IV) carbonate is shown below.

Sn(CO3)2(s)SnO2(s)+CO2(g)

The numbers of atoms of elements on the left side of the reaction are not equal to the right side of the reaction. Therefore, the reaction is unbalanced.

A coefficient of 2 is placed in front of CO2 to balance the chemical equation.

Therefore, the balanced equation for the decomposition reaction of solid tin (IV) carbonate is shown below.

Sn(CO3)2(s)SnO2(s)+2CO2(g)

The numbers of atoms of each element on the left side of the reaction are equal to the right side of the reaction. Therefore, the reaction is balanced.

Conclusion

The balanced equation for the decomposition reaction of solid tin (IV) carbonate is shown below.

Sn(CO3)2(s)SnO2(s)+2CO2(g)

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Chapter 7 Solutions

INTRODUCTORY CHEMISTRY-STD.GDE.+SOL.MAN

Ch. 7 - Prob. 11CECh. 7 - Prob. 12CECh. 7 - Prob. 13CECh. 7 - Prob. 14CECh. 7 - Prob. 15CECh. 7 - Prob. 16CECh. 7 - Prob. 1KTCh. 7 - Prob. 2KTCh. 7 - Prob. 3KTCh. 7 - Prob. 4KTCh. 7 - Prob. 5KTCh. 7 - Prob. 6KTCh. 7 - Prob. 7KTCh. 7 - Prob. 8KTCh. 7 - Prob. 9KTCh. 7 - Prob. 10KTCh. 7 - Prob. 11KTCh. 7 - Prob. 12KTCh. 7 - Prob. 13KTCh. 7 - Prob. 14KTCh. 7 - Prob. 15KTCh. 7 - Prob. 16KTCh. 7 - Prob. 17KTCh. 7 - Prob. 18KTCh. 7 - Prob. 19KTCh. 7 - Prob. 20KTCh. 7 - Prob. 21KTCh. 7 - Prob. 22KTCh. 7 - Prob. 1ECh. 7 - Prob. 2ECh. 7 - Prob. 3ECh. 7 - Prob. 4ECh. 7 - Prob. 5ECh. 7 - Prob. 6ECh. 7 - Prob. 7ECh. 7 - Prob. 8ECh. 7 - Prob. 9ECh. 7 - Prob. 10ECh. 7 - Prob. 11ECh. 7 - Prob. 12ECh. 7 - Prob. 13ECh. 7 - Prob. 14ECh. 7 - Prob. 15ECh. 7 - Prob. 16ECh. 7 - Prob. 17ECh. 7 - Prob. 18ECh. 7 - Prob. 19ECh. 7 - Prob. 20ECh. 7 - Prob. 21ECh. 7 - Prob. 22ECh. 7 - Prob. 23ECh. 7 - Prob. 24ECh. 7 - Prob. 25ECh. 7 - Prob. 26ECh. 7 - Prob. 27ECh. 7 - Prob. 28ECh. 7 - Prob. 29ECh. 7 - Prob. 30ECh. 7 - Prob. 31ECh. 7 - Prob. 32ECh. 7 - Prob. 33ECh. 7 - Prob. 34ECh. 7 - Prob. 35ECh. 7 - Prob. 36ECh. 7 - Prob. 37ECh. 7 - Prob. 38ECh. 7 - Prob. 39ECh. 7 - Prob. 40ECh. 7 - Prob. 41ECh. 7 - Prob. 42ECh. 7 - Prob. 43ECh. 7 - Prob. 44ECh. 7 - Prob. 45ECh. 7 - Prob. 46ECh. 7 - Prob. 47ECh. 7 - Prob. 48ECh. 7 - Prob. 49ECh. 7 - Prob. 50ECh. 7 - Prob. 51ECh. 7 - Prob. 52ECh. 7 - Prob. 53ECh. 7 - Prob. 54ECh. 7 - Prob. 55ECh. 7 - Prob. 56ECh. 7 - Prob. 57ECh. 7 - Prob. 58ECh. 7 - Prob. 59ECh. 7 - Prob. 60ECh. 7 - Prob. 61ECh. 7 - Prob. 62ECh. 7 - Prob. 63ECh. 7 - Prob. 64ECh. 7 - Prob. 65ECh. 7 - Prob. 66ECh. 7 - Prob. 67ECh. 7 - Prob. 68ECh. 7 - Prob. 69ECh. 7 - Prob. 70ECh. 7 - Prob. 71ECh. 7 - Prob. 72ECh. 7 - Prob. 73ECh. 7 - Prob. 74ECh. 7 - Prob. 75ECh. 7 - Prob. 76ECh. 7 - Prob. 77ECh. 7 - Prob. 78ECh. 7 - Prob. 79ECh. 7 - Prob. 80ECh. 7 - Prob. 81ECh. 7 - Prob. 82ECh. 7 - Prob. 83ECh. 7 - Prob. 84ECh. 7 - Prob. 85ECh. 7 - Prob. 86ECh. 7 - Prob. 87ECh. 7 - Prob. 88ECh. 7 - Prob. 1STCh. 7 - Prob. 2STCh. 7 - Prob. 3STCh. 7 - Prob. 4STCh. 7 - Prob. 5STCh. 7 - Prob. 6STCh. 7 - Prob. 7STCh. 7 - Prob. 8STCh. 7 - Prob. 9STCh. 7 - Prob. 10STCh. 7 - Prob. 11STCh. 7 - Prob. 12STCh. 7 - Prob. 13STCh. 7 - Prob. 14STCh. 7 - Prob. 15STCh. 7 - Prob. 16STCh. 7 - Prob. 17STCh. 7 - Prob. 18ST
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