Explanation Given information: The amount of petroleum burned is approx. 9 .0 × 10 12 kg/year All the petroleum is assumed to be in the form of octane. The existing amount of atmospheric CO 2 is, 3 .0 × 10 15 kg The equation for combustion of octane by using the chemical formulas for all reactants and products is as follows: C 8 H 18 ( g ) + O 2 ( g ) → CO 2 ( g ) +H 2 O ( l ) Balance the above equation by starting with hydrogen as follows: C 8 H 18 ( g ) + O 2 ( g ) → CO 2 ( g ) +H 2 O ( l ) 18 H atoms → 2 H atoms To balance hydrogen atoms, put 9 before H 2 O C 8 H 18 ( g ) + O 2 ( g ) → CO 2 ( g ) +9H 2 O ( l ) 18 H atoms → 18 H atoms Here, hydrogen atoms are balanced. Next, balance carbon as follows: C 8 H 18 ( g ) + O 2 ( g ) → CO 2 ( g ) +9H 2 O ( l ) 8 C atoms → 1 C atom To balance carbon atoms, put 8 before CO 2 C 8 H 18 ( g ) + O 2 ( g ) → 8 CO 2 ( g ) +9H 2 O ( l ) 8 C atoms → 8 C atoms Here, carbon atoms are balanced. Next, balance oxygen as follows: C 8 H 18 ( g ) + O 2 ( g ) → 8 CO 2 ( g ) +9H 2 O ( l ) 2 O atoms → 25 O atoms To balance oxygen atoms, put 25 2 before O 2 C 8 H 18 ( g ) + ( 25 / 2 ) O 2 ( g ) → 8 CO 2 ( g ) +9H 2 O ( l ) 25 O atoms → 25 O atoms Here, oxygen atoms are balanced. The balanced equation can be simplified by multiplying with 2 on both reactant and product sides as: 2C 8 H 18 ( g ) + 25 O 2 ( g ) → 16 CO 2 ( g ) +18H 2 O ( l ) Reactants Products 16 C atoms 16 C atoms 36 H atoms 36 H atoms 50 O atoms 50 O atoms Thus, the equation is balanced. From the equation it is clear that 2 mol C 8 H 18 reacts with 25 mol O 2 to form 16 mol CO 2 and 18 mol H 2 O

Introductory Chemistry (5th Edition) (Standalone Book)

5th Edition

ISBN: 9780321910295

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 8, Problem 101E

Interpretation Introduction

Interpretation:

The amount of CO2 produced by world fossil fuel combustion per year and the time to double the existing amount of atmospheric CO2 is to be calculated.

Concept Introduction: Conversion factors involving moles of reactants and products can be derived from a balanced chemical equation using the coefficients.

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Chapter 8, Problem 100E

Chapter 8, Problem 102E

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Introductory Chemistry (5th Edition) (Standalone Book)

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Solution Summary: The author explains how the conversion factors involving moles of reactants and products can be derived from a balanced chemical equation using the coefficients.

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