Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 5, Problem 159CP
Interpretation Introduction

Interpretation: The formula of hydrocarbon in the given reaction of combustion of this hydrocarbon with O2 at 0.959atm and 298K is needed to be identified, if the combustion producing a mixture of CO2 and H2O having a density of 1.391g/L and occupies a volume of 4 times larger than hydrocarbon.

Concept introduction:

  • Number of moles of a substance,
  • According to ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

  • Mass of a substance from its density and volume is,

Mass=Density×Volume

Mass of a substance from its number of moles is,

Number of moles×Molecularmass in grams=takenMass

  • Balance equation of a reaction is written according to law of conservation of mass.
  • Mole ratio between the reactants of a reaction is depends upon the coefficients of reactants in a balanced chemical equation.

Expert Solution & Answer
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Answer to Problem 159CP

Answer

The formula of hydrocarbon in the given reaction is C2H6 .

Explanation of Solution

Explanation

  • To find: the balanced chemical equation of the given reaction of combustion of hydrocarbon compound.

CxHy+O2xCO2+y2H2O

Let’s take the hydrocarbon as CxHy .

The combustion of CxHy in O2 is producing CO2 and H2O gases.

Therefore,

The chemical equation of the reaction given is,

CxHy+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 and then oxygen atoms.

Therefore, the balanced chemical equation of the given reaction of combustion of hydrocarbon compound will be,

CxHy+O2xCO2+y2H2O

Hence,

The balanced chemical equation for the given reaction of combustion of hydrocarbon compound is, CxHy+O2xCO2+y2H2O .

The volume of hydrocarbon is taken as 1 L.

To find: the number of moles of reacted CxHy in the given reaction.

The number of moles of CxHy is 0.0392mol .

The pressure of reacted CxHy in the given reaction is given as 0.959atm .

The volume of CxHy is 1L .

The temperature is given as 298K .

According to ideal gas equation is,

Numberofmoles=Pressure×VolumeR×Temperature

Therefore, the number of moles of CxHy ,

nCxHy=0.959atm×1L0.08206Latm/Kmol×298K=0.0392mol

Hence, the number of moles of CxHy is in given reaction is 0.0392mol .

The volume of mixture of products is taken as 4 L (since the volume of hydrocarbon is taken as 1 L).

To find: the number of moles and mass of produced mixture of CO2 and H2O having a density of 1.391g/L in the given reaction.

The number of moles of produced mixture is 0.196mol .

The mass of produced mixture is 5.564g .

The pressure of produced mixture in the given reaction is given as 1.51atm .

The volume of produced mixture is 4L .

The temperature is given as 375K .

According to ideal gas equation is,

Numberofmoles=Pressure×VolumeR×Temperature

Therefore, the number of moles of produced mixture is,

nmixture=1.51atm×4L0.08206Latm/Kmol×375K=0.196mol

Hence, the number of moles of produced mixture is in given reaction is 0.196mol .

The density of produced mixture is given as 1.391g/L

Mass of a substance from its density and volume is,

Mass=Density×Volume

Therefore,

The mass of produced mixture is,

Mass=1.391g/L×4L=5.564g

To find: the number of moles of CO2 and H2O in the produced mixture of given reaction.

The number of moles of CO2 reacted is 2×0.0392mol .

The number of moles of H2O produced is 4×0.0392mol

  • The balanced chemical equation for the given reaction of combustion of hydrocarbon compound is, CxHy+O2xCO2+y2H2O .

Here, the 1 mole of CxHy is producing xmolCO2andy2molH2O .

The number of moles of CxHy is in given reaction is calculated as 0.0392mol .

Therefore,

0.0392mol of CxHy is producing (0.0392)xmolCO2and(0.0392)y2molH2O .

The number of moles and mass of produced mixture of CO2 and H2O is calculated as 0.196mol .

That means,

(0.0392)xmol+(0.0392)y2mol=0.196mol -Equation_1

  • The mass of produced mixture of CO2 and H2O is calculated as 5.564g .

The molecular mass of CO2 and H2O are 44g and 18g respectively.

Mass of a substance from its number of moles is,

Number of moles×Molecularmass in grams=takenMass

Therefore,

(0.0392x)(44)+(0.0392y2)(18)=5.564g -Equation_2

To solve the both equations,

Multiply the equation_1 with (44) and subtract with equation­_2.

That is,

[(0.0392x)(44)+(0.0392)y2(44)=44×0.196][(0.0392x)(44)+(0.0392y2)(18)=5.564g]=(0.0392y2)(26)=8.6245.5640.0392y2=326 y=3×226×0.0392=6

The value of y is found that 6, so the value of x will be 2 (since (0.0392)xmol+(0.0392)y2mol=0.196mol -Equation_1 ).

To determine: the formula of hydrocarbon ( CxHy ) in the given reaction of combustion of this hydrocarbon with O2 at 0.959atm .

The formula of hydrocarbon in the given reaction is C2H6 .

The general formula of the hydrocarbon is CxHy .

The values of x and y are calculated as 2 and 6.

Therefore, the formula of hydrocarbon in the given reaction is C2H6 .

Conclusion

Conclusion

The formula of the reactant hydrocarbon of given reaction is identified. The formula of hydrocarbon ( CxHy ) contains C element and H element, so it is depend upon the number of moles of produced CO2 and H2O in the given reaction. The number of moles of these molecules is calculated from their given data’s.

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Chapter 5 Solutions

Chemistry

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