Chemistry for Engineering Students
Chemistry for Engineering Students
3rd Edition
ISBN: 9781285199023
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Textbook Question
Chapter 4, Problem 4.87PAE

4.93 A mixture of methane (CH4) and propane (C3H8) has a total mass of 29.84 g. When the mixture is burned completely in excess oxygen, the CO2 and H2O products have a combined mass of 142.97 g. Calculate the mass of methane in the original mixture.

Expert Solution & Answer
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Interpretation Introduction

Interpretation: The mass of methane in the original mixture should be found out if CO2 and H2O products result in a combined mass of 142.97 g when a mixture of methane ( CH4 ) and propane ( C3H8 ) with a total mass of 29.84 g is burned completely in the presence of excess oxygen.

Concept introduction:

Burning of methane and propane in excess oxygen is a combustion reaction. When hydrocarbons burn out, that results in water and carbon dioxide gas. To complete combustion, there should be enough oxygen supply. Otherwise half combustion can occur. As a result of half combustion carbon monoxide and carbon particles are released to the environment.

Answer to Problem 4.87PAE

Solution:

Mass of methane in original sample is12.53 g 

Explanation of Solution

Completecombustion (Balance equation) reaction for both CH4 and C3H8 can be written as follows.

CH4+ 2O2 CO2+ 2H2O

C3H8+ 5O23CO2+ 4H2O

Let’s take initial weight of propane as X, then initial weight of methane should be (29.84-X).

n = mM (n = no of moles, m = weight of the sample, M = molar mass)

Using above equation, number of moles of each molecules present can be found out.

In initial sample,Number of moles of methane=(29.84X)16Numberofmoleso

Nunber of moles of propane= X44

After the combustion of methane,

Number of moles of CO2formed= (29.84X)16Number of moles of H2O formed= 2× (29.84X)16

After the combustion of propane,

Number of moles of CO2formed=3× X44Number of moles of H2O   = 4×X44

So final products weight can be obtained as follows,

(Molar mass of CO2×Total number of moles of CO2)+(Molar mass of H2O×Total number of moles of H2O)=142.9744×( (29.84X)16+ 3x44)     +   18×(2×(29.84X)16 + 4X44  ) = 142.97 X=17.31

So the mass of methane in original sample = 29.84 g -17.31 g

                            = 12.53 g

Conclusion

When the mas of a hydrocarbon mixture is known along with the mass of the product mix formed when it has undergone combustion reaction, the mass of individual hydrocarbons can be found out if the corresponding molar masses of the hydrocarbons are known.

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Chapter 4 Solutions

Chemistry for Engineering Students

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