Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Textbook Question
Chapter 8, Problem 155CP

Methane (CH4) gas flows into a combustion chamber at a rate of 200. L/min at 1.50 atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited.

a. To ensure complete combustion of CH4 to CO2(g) and H2O(g), three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent O2 and 79 mole percent N2, calculate the flow rate of air necessary to deliver the required amount of oxygen.

b. Under the conditions in part a, combustion of methane was not complete as a mixture of CO2(g) and CO(g) was produced. It was determined that 95.0% of the carbon in the exhaust gas was present in CO2. The remainder was present as carbon in CO. Calculate the composition of the exhaust gas in terms of mole fraction of CO, CO2, O2, N2, and H2O. Assume CH4 is completely reacted and N2 is unreacted.

a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The flow rate of air is needed to complete the given reaction of combustion of CH4 if the combustion of CH4 is producing COandCO2 are needed to be determined.

Concept introduction:

  • Balanced equation of a reaction is written according to law of conservation of mass.
  • According to ideal gas law: the pressure and volume of a gas is directly proportional to the number of moles of gas at constant temperature.

PVn=RT, P,V and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

P1V1n1=P2V2n2

  • Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.
  • Equation for mole fraction of a molecule in a mixture of molecules is,
  • moleculefractionofA,(χA=numbersof moles ofmolecule A(nA)total number of moles(ntotal

  • Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.
  • Total pressure of a flask containing mixture of gases is the sum of individual partial pressures of constituted gases.

To determine: the flow rate of air needed to complete the given reaction of combustion of CH4 if the combustion of CH4 is producing COandCO2.

Answer to Problem 155CP

The flow rate of air needed to complete the given reaction of combustion of CH4 is 8.7×103L/min.

Explanation of Solution

To find: the balanced chemical equation of the given reaction.

CH4+2O2CO2+2H2O

The reactants of the given reaction are CH4 and O2.

The products of the given reaction are CO2 and H2O.

Therefore,

The chemical equation of the reaction given is,

CH4+O2CO2+H2O

This equation is not balanced, Balanced equation is the chemical equation of a reaction which is written according to law of conservation of mass the mass of reactant molecules should conserve in product molecules.

The balanced chemical equation of the given reaction is,

Balance the molecules in both sides of arrow in order of atoms other than hydrogen, hydrogen atoms then oxygen atoms.

Therefore, the balanced chemical equation of the given reaction will be,

CH4+2O2CO2+2H2O

Hence,

The balanced chemical equation for the given reaction is, CH4+2O2CO2+2H2O.

Assumes the number of moles of CH4 is as x.

To find: the number of moles of required O2 and air in the given reaction.

The number of moles of required O2 is 6xmol.

The number of moles of required air is 29xmol.

The balanced chemical equation for the given reaction is, CH4+2O2CO2+2H2O.

Here, the combustion of 1 mole of CH4 requires 2 mole of O2.

But to ensure the completion of combustion of CH4, three times as much oxygen is necessary is required. Therefore,

If the number of moles of CH4 is as x, then the number of moles of required O2 is,

3×xmolCH4×2molO21molCH4=6xmolO2

The mole percent of O2 in the air is 21.

That means,

nO2=0.21nair

Therefore,

The number of moles of required air is,

nair=6xmol0.21=29xmol

Hence,

The number of moles of required O2 is 6xmol.

The number of moles of required air is 29xmol.

To determine: the flow rate of air needed to complete the given reaction of combustion of CH4.

The flow rate of air needed to complete the given reaction of combustion of CH4 is 8.7×103L/min.

The volume of CH4 is given as 200L/min.

The pressure of CH4 is given as 1.50atm.

The number of moles of CH4 is x.

The pressure of air is given as 1atm.

The number of moles of required air is calculated as 29xmol.

According to ideal gas law, the pressure and volume of a gas is directly proportional to the number of moles of gas at constant temperature.

PVn=RT, P,V and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

P1V1n1=P2V2n2

Take gas-1 as CH4 and gas-2 as air,

Therefore,

The volume of air is,

200L/min×1.50atm×29xmol1atm×xmol=8.7×103L/min

Hence,

The flow rate of air needed to complete the given reaction of combustion of CH4 is 8.7×103L/min.

b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The composition of the exhaust gas in terms of mole fraction of CO,CO2,O2,N2andH2O if the combustion of CH4 is producing COandCO2 are needed to be determined.

Concept introduction:

  • Balanced equation of a reaction is written according to law of conservation of mass.
  • According to ideal gas law: the pressure and volume of a gas is directly proportional to the number of moles of gas at constant temperature.

PVn=RT, P,V and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

P1V1n1=P2V2n2

  • Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.
  • Equation for mole fraction of a molecule in a mixture of molecules is,
  • moleculefractionofA,(χA=numbersof moles ofmolecule A(nA)total number of moles(ntotal

  • Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.
  • Total pressure of a flask containing mixture of gases is the sum of individual partial pressures of constituted gases.

To determine: the composition of the exhaust gas in terms of mole fraction of CO,CO2,O2,N2andH2O if the    combustion of CH4 is producing COandCO2.

Answer to Problem 155CP

The mole fraction of CO in the exhaust gas of given reaction is 0.0017.

The mole fraction of CO2 in the exhaust gas of given reaction is 0.0032.

The mole fraction of O2 in the exhaust gas of given reaction is 0.13.

The mole fraction of N2 in the exhaust gas of given reaction is 0.77.

The mole fraction of H2O in the exhaust gas of given reaction is 0.067.

Explanation of Solution

To find: the balanced chemical equation of the given reactions.

CH4+2O2CO2+2H2O

CH4+32O2CO+2H2O

The reactants of the given reactions are CH4 and O2.

The products of the given reaction are CO, CO2 and H2O.

Therefore,

The chemical equations of the given reactions are,

CH4+O2CO2+H2O

CH4+O2CO+H2O

These equations are not balanced, Balanced equation is the chemical equation of a reaction which is written according to law of conservation of mass the mass of reactant molecules should conserve in product molecules.

The balanced chemical equations of given reactions are,

Balance the molecules in both sides of arrow in order of atoms other than hydrogen, hydrogen atoms then oxygen atoms.

Therefore, the balanced chemical equation of the given reactions will be,

CH4+2O2CO2+2H2O

CH4+32O2CO+2H2O

Hence,

The balanced chemical equations for the given reactions are,

CH4+2O2CO2+2H2O

CH4+32O2CO+2H2O

#assumes the number of moles of CH4 is as x.

To find: the number of moles of produced CO, CO2, N2,H2O and remaining O2 in the given reactions.

The number of moles of produced CO in the given reaction is 0.05xmol.

The number of moles of produced CO2 in the given reaction is 0.95xmol.

The number of moles of N2 in the given reaction is 23xmol.

The number of moles of produced H2O in the given reactions is 2xmol.

The number of moles remaining O2 in the given reactions is 4.03xmol.

  • The carbon containing reactant is only CH4, but the carbon containing products are CO2 and CO.

    The produced CO2 containing 95% of carbon in the reactant CH4.

    The number of moles of CH4 is taken as x mole.

    Therefore,

    The number of moles of producedCO2 in the given reaction is 0.95xmol.

    So the remaining 5%  of carbon is contained by CO.

    Therefore,

    The number of moles of producedCO in the given reaction is 0.05xmol.

  • The number of moles of required air in the given reaction is calculated as 29xmol.

    The 79 mole percent of air is N2.

    That means,

    nN2=0.79nair=0.79×29xmol23xmol

    The number of moles ofN2 in the given reaction is 23xmol.

  • The number of moles of CH4 is taken as x mole.

    The balanced chemical equations for the given reactions are,

CH4+2O2CO2+2H2O

CH4+32O2CO+2H2O

Therefore,

The number of moles of produced H2O in the given reactions is,

2×xmolCH4×2molH2O2×1molCH4=2xmolH2O

The number of moles of producedH2O in the given reactions is 2xmol.

  • The number of moles of produced CO in the given reaction is 0.05xmol.

    The number of moles of produced CO2 in the given reaction is 0.95xmol.

    The reacted O2 in the given reaction is,

    0.95xmolCO2×2molO21molCO2=1.9xmolO2

    0.05xmolCO×1.5molO21molCO=0.075xmolO2

    That is,

    1.9xmol+0.075xmol=1.975xmol.

  • The number of moles of required O2 in the given reaction is calculated as 6xmol.

    The reacted O2 in the given reaction is 1.975xmol.

    Therefore,

    The number of moles remainingO2 in the given reactions is,

    6xmol1.975xmol4.03xmol

To determine: the compositions of the exhaust gas in terms of mole fraction of CO,CO2,O2,N2andH2O if the    combustion of CH4 is producing COandCO2.

The mole fraction of CO in the exhaust gas of given reaction is 0.0017.

The mole fraction of CO2 in the exhaust gas of given reaction is 0.0032.

The mole fraction of O2 in the exhaust gas of given reaction is 0.13.

The mole fraction of N2 in the exhaust gas of given reaction is 0.77.

The mole fraction of H2O in the exhaust gas of given reaction is 0.067.

The number of moles of produced CO in the given reaction is calculated as 0.05xmol.

The number of moles of produced CO2 in the given reaction is calculated as 0.95xmol.

The number of moles of N2 in the given reaction is calculated as 23xmol.

The number of moles of produced H2O in the given reactions is calculated as 2xmol.

The number of moles remaining O2 in the given reactions is calculated as 4.03xmol.

These molecules are presented in the exhaust gas.

Therefore,

The total number of moles of in the exhaust gas is,

0.05xmol+0.95xmol+23xmol+2xmol+4.03xmol=30xmol.

Equation for mole fraction of a molecule in a mixture of molecules is,

moleculefractionofA,(χA=numbersof moles ofmolecule A(nA)total number of moles(ntotal

Therefore,

The mole fraction of CO in the exhaust gas is,

0.05xmol30xmol=0.0017

The mole fraction of CO2 in the exhaust gas is,

0.95xmol30xmol=0.032

The mole fraction of N2 in the exhaust gas of given reaction is

23xmol30xmol=0.77

The mole fraction of H2O in the exhaust gas of given reaction is

2xmol30xmol=0.067

The mole fraction of O2 in the exhaust gas is,

4.03xmol30xmol=0.13

Hence,

The mole fraction of CO in the exhaust gas of given reaction is 0.0017.

The mole fraction of CO2 in the exhaust gas of given reaction is 0.0032.

The mole fraction of O2 in the exhaust gas of given reaction is 0.13.

The mole fraction of N2 in the exhaust gas of given reaction is 0.77.

The mole fraction of H2O in the exhaust gas of given reaction is 0.067.

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Chapter 8 Solutions

Chemistry: An Atoms First Approach

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