At the end of this, how did we go from a 5 to 1 ratio moles to 2CO2 and 3H2O?

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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At the end of this, how did we go from a 5 to 1 ratio moles to 2CO2 and 3H2O?

Number of moles of hydrocarbon is calculated as follows:
PV = nRT
Rearrange above equation for n.
PV = nRT
n = RT
Value of Vis V.
Value of R is 0.082057 L atm K-¹.mol-¹.
Value of Tis 298 K.
Value of P is 0.959 atm.
Substitute the values in above equation for moles of hydrocarbon.
n = RT
(0.959 atm) (V)
(0.082057 L-atm-K-¹-mol-¹) (298 K)
= 0.0392 V mol
Hence, number of moles of hydrocarbon is 0.0392 V mol.
Value of Vis 4V.
Value of R is 0.082057 L atm K-¹ mol-¹.
Value of Tis 375 K.
Value of P is 1.51 atm.
Substitute the values in above equation for moles of mixture of gases.
n = RT
(1.51 atm)(4V)
(0.082057 L-atm-K-¹-mol-¹)(375 K)
= 0.1962 V mol
Hence, number of moles of carbon dioxide and water vapor mixture is 0.1962 V mol.
Ratio of number of moles of hydrocarbons to mixture of gases is as follows:
number of moles of hydrocarbon
Ratio =
number of moles of mixture
Value of number of moles of hydrocarbon is 0.0392 V mol.
Value of number of moles of mixture is 0.1962 V mol.
Substitute values in above expression.
Ratio =
number of moles of mixture
number of moles of hydrocarbon
=
Conclusion
0.1962 V mol
0.0392 V mol
Hence, total number of moles of mixture of carbon dioxide and water vapor is 5 and hydrocarbon is 1.
Therefore, combustion reaction of hydrocarbon so formed is as follows:
Hydrocarbon + O2 → 2CO2 + 3H₂O
From above reaction it is evident that number of carbon and hydrogen in molecular formula of hydrocarbon
is 2 and 6. Hence, molecular formula of hydrocarbon is C₂H6.
Molecular formula of hydrocarbon is C₂H6.
Transcribed Image Text:Number of moles of hydrocarbon is calculated as follows: PV = nRT Rearrange above equation for n. PV = nRT n = RT Value of Vis V. Value of R is 0.082057 L atm K-¹.mol-¹. Value of Tis 298 K. Value of P is 0.959 atm. Substitute the values in above equation for moles of hydrocarbon. n = RT (0.959 atm) (V) (0.082057 L-atm-K-¹-mol-¹) (298 K) = 0.0392 V mol Hence, number of moles of hydrocarbon is 0.0392 V mol. Value of Vis 4V. Value of R is 0.082057 L atm K-¹ mol-¹. Value of Tis 375 K. Value of P is 1.51 atm. Substitute the values in above equation for moles of mixture of gases. n = RT (1.51 atm)(4V) (0.082057 L-atm-K-¹-mol-¹)(375 K) = 0.1962 V mol Hence, number of moles of carbon dioxide and water vapor mixture is 0.1962 V mol. Ratio of number of moles of hydrocarbons to mixture of gases is as follows: number of moles of hydrocarbon Ratio = number of moles of mixture Value of number of moles of hydrocarbon is 0.0392 V mol. Value of number of moles of mixture is 0.1962 V mol. Substitute values in above expression. Ratio = number of moles of mixture number of moles of hydrocarbon = Conclusion 0.1962 V mol 0.0392 V mol Hence, total number of moles of mixture of carbon dioxide and water vapor is 5 and hydrocarbon is 1. Therefore, combustion reaction of hydrocarbon so formed is as follows: Hydrocarbon + O2 → 2CO2 + 3H₂O From above reaction it is evident that number of carbon and hydrogen in molecular formula of hydrocarbon is 2 and 6. Hence, molecular formula of hydrocarbon is C₂H6. Molecular formula of hydrocarbon is C₂H6.
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