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

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

Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter16: Spontaneity, Entropy, And Free Energy
Section: Chapter Questions
Problem 121MP: Consider a sample containing 5.00 moles of a monatomic ideal gas that is taken from state A to state...
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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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