Chemistry: An Atoms-Focused Approach (Second Edition)
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 7, Problem 7.23QA
Interpretation Introduction

To write:

The complete and balanced chemical equation for the given incomplete reactions.

Expert Solution & Answer
Check Mark

Answer to Problem 7.23QA

Solution:

a. C3H8  g+ 5O2  g  3CO2  g+ 4H2Og

b. 2C4H10  g+ 13O2  g  8CO2  g+ 10H2Og

c. 2C6H6  l+ 15O2  g  12CO2  g+ 6H2Og

d. 2C8H18  l+ 25O2  g  16CO2  g+ 18H2Og

Explanation of Solution

First, we write the complete and balanced chemical equation describing complete combustion. Combustion involves a rapid reaction with O2. The products of complete combustion of hydrocarbons are CO2 and H2O.The first step involves writing a reaction expression with single molecules or formula unit of the reactants and products. Then we take the inventories of the number of atoms of each of the elements in the reaction mixture and we balance them starting with those that appear in only one reactant and product.

a. C3H8  g+ O2  g  CO2  g+ H2Og

Atoms: 3C+8H+2O 1C+3O+ 2H

There is only one C atom on the right and three on left hand side of reaction arrow. To balance C, we place a coefficient of 3 in front of CO2 and recalculate the distribution of atoms on both sides.

C3H8  g+ O2  g  3CO2  g+ H2Og

Atoms: 3C+8H+2O 3C+7O+ 2H

There are two H atoms on the right and eight on left hand side of reaction arrow. To balance H,

we place a coefficient of 4 in front of H2O and recalculate distribution of atoms on both sides.

C3H8  g+ O2  g  3CO2  g+ 4H2Og

Atoms: 3C+8H+2O 3C+10O+ 8H

There are ten O atoms on the right and two on left hand side of reaction arrow. To balance O, we place a coefficient of 5 in front of O2 and recalculate distribution of atoms on both sides.

C3H8  g+ 5O2  g  3CO2  g+ 4H2Og

Atoms: 3C+8H+10O 3C+10O+ 8H

Both sides of reaction have the same number of each element. Therefore, this reaction is balanced.

b. C4H10  g+ O2  g  CO2  g+ H2Og

Atoms: 4C+10H+2O 1C+3O+ 2H

There is only one C atom on the right and four on left hand side of reaction arrow. To balance  , we place a coefficient of 4 in front of CO2 and recalculate distribution of atoms on both sides.

C4H10  g+ O2  g  4CO2  g+ H2Og

Atoms: 4C+10H+2O 4C+9O+ 2H

There are two H atoms on the right and ten on left hand side of reaction arrow. To balance H, we place a coefficient of 5 in front of H2O and recalculate distribution of atoms on both sides.

C4H10  g+ O2  g  4CO2  g+ 5H2Og

Atoms: 4C+10H+2O 4C+13O+ 10H

This balances the number of C and H atoms but leaves an odd number of O atoms on the product side. The only source of O atoms is O2, so we need an even number of O atoms on the right. To balance O atoms, multiply all the coefficients in the expression by 2.

2C4H10  g+ 2O2  g  8CO2  g+ 10H2Og

Atoms: 8C+20H+4O 8C+26O+ 20H

We can now balance the number of O atoms by replacing the coefficient 2 in front of O2 with 13, which gives us 26 O atoms and a balanced equation.

2C4H10  g+ 13O2  g  8CO2  g+ 10H2Og

Atoms: 8C+20H+26O 8C+26O+ 20H

c. C6H6  l+ O2  g  CO2  g+ H2Og

Atoms: 6C+6H+2O 1C+3O+ 2H

There is only one C atom on the right and six on left hand side of reaction arrow. To balance C , we place a coefficient of 6 in front of CO2 and recalculate distribution of atoms on both sides.

C6H6  l+ O2  g  6CO2  g+ H2Og

Atoms: 6C+6H+2O 6C+13O+ 2H

There are two H atoms on the right and six on left hand side of reaction arrow. To balance H

 we place a coefficient of 3 in front of H2O and recalculate distribution of atoms on both sides.

C6H6  l+ O2  g  6CO2  g+ 3H2Og

Atoms: 6C+6H+2O 6C+15O+ 6H

This balances the number of C and H atoms but leaves an odd number of O atoms on the product side. The only source of O atoms is O2, so we need an even number of O atoms on the right. To balance O atoms, multiply all the coefficients in the expression by 2.

2C6H6  l+ 2O2  g  12CO2  g+ 6H2Og

Atoms: 12C+12H+4O 12C+30O+ 12H

We can now balance the number of O atoms by replacing the coefficient 2 in front of O2 with 15, which gives us 30 O atoms and a balanced equation.

2C6H6  l+ 15O2  g  12CO2  g+ 6H2Og

Atoms: 12C+12H+30O 12C+30O+ 12H

d. C8H18  l+ O2  g  CO2  g+ H2Og

Atoms: 8C+18H+2O 1C+3O+ 2H

There is only one C atom on the right and eight on left hand side of reaction arrow. To balance C, we place a coefficient of 8 in front of CO2 and recalculate distribution of atoms on both sides.

C8H18  l+ O2  g  8CO2  g+ H2Og

Atoms: 8C+18H+2O 8C+17O+ 2H

There are two H atoms on the right and 18 on left hand side of reaction arrow. To balance H

we place a coefficient of 9 in front of H2O and recalculate distribution of atoms on both sides.

C8H18  l+ O2  g  8CO2  g+ 9H2Og

Atoms: 8C+18H+2O 8C+25O+ 18H

This balances the number of C and H atoms but leaves an odd number of O atoms on the product side. The only source of O atoms is O2, so we need an even number of O atoms on the right. To balance O atoms, multiply all the coefficients in the expression by 2.

2C8H18  l+ 2O2  g  16CO2  g+ 18H2Og

Atoms: 16C+36H+4O 16C+50O+ 36H

We can now balance the number of O atoms by replacing the coefficient 2 in front of O2 with 25, which gives us 30 O atoms and a balanced equation.

2C8H18  l+ 25O2  g  16CO2  g+ 18H2Og

Atoms: 16C+36H+50O 16C+50O+ 36H

Conclusion:

Balancing each atom on both sides of reaction arrow balances the entire reaction.

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Chapter 7 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

Ch. 7 - Prob. 7.11QACh. 7 - Prob. 7.12QACh. 7 - Prob. 7.13QACh. 7 - Prob. 7.14QACh. 7 - Prob. 7.15QACh. 7 - Prob. 7.16QACh. 7 - Prob. 7.17QACh. 7 - Prob. 7.18QACh. 7 - Prob. 7.19QACh. 7 - Prob. 7.20QACh. 7 - Prob. 7.21QACh. 7 - Prob. 7.22QACh. 7 - Prob. 7.23QACh. 7 - Prob. 7.24QACh. 7 - Prob. 7.25QACh. 7 - Prob. 7.26QACh. 7 - Prob. 7.27QACh. 7 - Prob. 7.28QACh. 7 - Prob. 7.29QACh. 7 - Prob. 7.30QACh. 7 - Prob. 7.31QACh. 7 - Prob. 7.32QACh. 7 - Prob. 7.33QACh. 7 - Prob. 7.34QACh. 7 - Prob. 7.35QACh. 7 - Prob. 7.36QACh. 7 - Prob. 7.37QACh. 7 - Prob. 7.38QACh. 7 - Prob. 7.39QACh. 7 - Prob. 7.40QACh. 7 - Prob. 7.41QACh. 7 - Prob. 7.42QACh. 7 - Prob. 7.43QACh. 7 - Prob. 7.44QACh. 7 - Prob. 7.45QACh. 7 - Prob. 7.46QACh. 7 - Prob. 7.47QACh. 7 - Prob. 7.48QACh. 7 - Prob. 7.49QACh. 7 - Prob. 7.50QACh. 7 - Prob. 7.51QACh. 7 - Prob. 7.52QACh. 7 - Prob. 7.53QACh. 7 - Prob. 7.54QACh. 7 - Prob. 7.55QACh. 7 - Prob. 7.56QACh. 7 - Prob. 7.57QACh. 7 - Prob. 7.58QACh. 7 - Prob. 7.59QACh. 7 - Prob. 7.60QACh. 7 - Prob. 7.61QACh. 7 - Prob. 7.62QACh. 7 - Prob. 7.63QACh. 7 - Prob. 7.64QACh. 7 - Prob. 7.65QACh. 7 - Prob. 7.66QACh. 7 - Prob. 7.67QACh. 7 - Prob. 7.68QACh. 7 - Prob. 7.69QACh. 7 - Prob. 7.70QACh. 7 - Prob. 7.71QACh. 7 - Prob. 7.72QACh. 7 - Prob. 7.73QACh. 7 - Prob. 7.74QACh. 7 - Prob. 7.75QACh. 7 - Prob. 7.76QACh. 7 - Prob. 7.77QACh. 7 - Prob. 7.78QACh. 7 - Prob. 7.79QACh. 7 - Prob. 7.80QACh. 7 - Prob. 7.81QACh. 7 - Prob. 7.82QACh. 7 - Prob. 7.83QACh. 7 - Prob. 7.84QACh. 7 - Prob. 7.85QACh. 7 - Prob. 7.86QACh. 7 - Prob. 7.87QACh. 7 - Prob. 7.88QACh. 7 - Prob. 7.89QACh. 7 - Prob. 7.90QACh. 7 - Prob. 7.91QACh. 7 - Prob. 7.92QACh. 7 - Prob. 7.93QACh. 7 - Prob. 7.94QACh. 7 - Prob. 7.95QACh. 7 - Prob. 7.96QACh. 7 - Prob. 7.97QACh. 7 - Prob. 7.98QACh. 7 - Prob. 7.99QACh. 7 - Prob. 7.100QACh. 7 - Prob. 7.101QACh. 7 - Prob. 7.102QACh. 7 - Prob. 7.103QACh. 7 - Prob. 7.104QACh. 7 - Prob. 7.105QACh. 7 - Prob. 7.106QACh. 7 - Prob. 7.107QACh. 7 - Prob. 7.108QACh. 7 - Prob. 7.109QACh. 7 - Prob. 7.110QA
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