Chapter 9, Problem 40QAP

### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

Chapter
Section

### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
81 views

# In the “Chemistry in Focus” segment Cars of the Future, the claim is made that the combustion of gasoline for some cars causes about 1 lb of CO 2 to be produced for each mile traveled. Estimate the gas mileage of a car that produces about 1 lb of CO 2 per mile traveled. Assume gasoline has a density of 0.75 g/mL and is 100% octane ( C 8 H 18 ) . While this last part is not true, it is close enough for an estimation. The reaction can be represented by the following unbalanced chemical equation::math> C 8 H 18 + O 2 → CO 2 + H 2 O

Interpretation Introduction

Interpretation:

The mileage of a car that produces 1lb of CO2 per mile should be calculated.

Concept Introduction:

The mileage of car is miles per gallon of gasoline used. The number of moles is related to mass and molar mass as follows:

n=mM

Here, m is mass and M is molar mass.

The following conversion factor should be used to perform the given problem.

1 mL=0.000264172 gallon

Also,

1 lb=453.6 g.

Explanation

The balanced chemical reaction for combustion of octane (gasoline) is as follows:

2C8H18+25O216CO2+18H2O

From the balanced chemical reaction, 16 mol of carbon dioxide obtained from 2 mol of octane. Thus, 1 mol of carbon dioxide is obtained from 216 mol of octane.

Mass of carbon dioxide produced is 1 lb or 453.6 g. Molar mass of carbon dioxide is 44.01 g/mol thus, number of moles will be:

n=mM

Putting the values,

n=453.6 g44.01 g/mol=10.31 mol

Now, 10.31 mol of carbon dioxide is obtained from 10.31×0.125=1.288 mol of octane.

Molar mass of octane is 114.23 g/mol thus, mass can be calculated as follows:

m=n×M=1

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