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Topic: Partial Pressure and mole fractions  (Read 7656 times)

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Offline trinhn812

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Partial Pressure and mole fractions
« on: January 07, 2008, 11:35:58 PM »
Natural gas is a mixture of hydrocarbons, primarily methane (CH4) and ethane (C2H6).  A typical mixture might have Χmethane = 0.915 and Χethane = 0.085. 

a.   What are the partial pressures of the two gases in a 15.00-L container of natural gas at 20.°C and 1.44 atm? 

b.   Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?

For part A is the total pressure given already, 1.44atm? So I would just multiplay the mole fraction times 1.44 giving partial pressures of (1.44*.915) = 1.32 atm for methane and (1.44*.085)=.1224 atm for ethane?

and for part B what might the equation be? CH4 + C2H6 + O2 = ?

Thanks for all your help.

Offline LQ43

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Re: Partial Pressure and mole fractions
« Reply #1 on: January 08, 2008, 04:32:56 PM »
Natural gas is a mixture of hydrocarbons, primarily methane (CH4) and ethane (C2H6).  A typical mixture might have Χmethane = 0.915 and Χethane = 0.085. 

a.   What are the partial pressures of the two gases in a 15.00-L container of natural gas at 20.°C and 1.44 atm? 

For part A is the total pressure given already, 1.44atm? So I would just multiplay the mole fraction times 1.44 giving partial pressures of (1.44*.915) = 1.32 atm for methane and (1.44*.085)=.1224 atm for ethane?


yes this is correct

b.   Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?


and for part B what might the equation be? CH4 + C2H6 + O2 = ?

Thanks for all your help.

 - make 2 equations for the 2 gases and balance them

CH4 + O2 -->  H2O + ?

C2H6 +  O2 -->  H2O  +  ?

 - you need to find total moles (how?) and use mole fraction to find moles of each gas

Then do the stoichiometry to find the mass of the water for each then add to get total mass of water

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