   # Camping stoves are fueled by propane (C 3 H 8 ), butane [C 4 H 10 (g), Δ f H ° = –127.1 kJ/mol]. gasoline. or ethanol (C 2 H 5 OH). Calculate the enthalpy of combustion per gram of each of these fuels. [Assume that gasoline is represented by isooctane, C 8 H 18 ( ℓ ) , with Δ f H ° = –259.3 kJ/mol.) Do you notice any great differences among these fuels? How are these differences related to their composition? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 5, Problem 84GQ
Textbook Problem
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## Camping stoves are fueled by propane (C3H8), butane [C4H10(g), ΔfH° = –127.1 kJ/mol]. gasoline. or ethanol (C2H5OH). Calculate the enthalpy of combustion per gram of each of these fuels. [Assume that gasoline is represented by isooctane, C8H18(ℓ), with ΔfH° = –259.3 kJ/mol.) Do you notice any great differences among these fuels? How are these differences related to their composition? Interpretation Introduction

Interpretation:

The enthalpy of combustion per gram of propane and butane has to be calculated

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state

ΔrH0=ΣnΔfH0(products)-ΣnΔfH0(reactants)

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

ΔrH=enthalpy changenumber of moles

ΔH=ΔrH×number of moles

### Explanation of Solution

For propane the balanced equation is as follows:

C3H8+ 5 O23CO2+ 4 H2O

ΔrH0=[(3mol×-393.5kJ/mol)+(4mol×-285.83kJ/mol)]-[(1mol×-104.7 kJ/mol+0)]

ΔrH0=-2219.12kJ

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

ΔrH=enthalpy changenumber of moles

ΔrH=-2219.12kJ/mol×1mol44.1g=-50.32J/g.K

So, the enthalpy of combustion is -50.32J/g

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