Natural gas is a mixture of hydrocarbons, primarily methane ( C H 4 ) and ethane

Question

Natural gas is a mixture of hydrocarbons, primarily methane (CH4) and ethane (C2H6). A typical mixture might have Xmethane = 0.915 and Xethane = 0.085. Let's assume that we have a 15.55g sample of natural gas in a volume of 11.70L at a temperature of 19.95?C.

How many total moles of gas are in the sample?

What is the pressure of the sample (in atmospheres)?

What is the partial pressure of each component in the sample (in atmospheres)?

When the sample is burned in an excess of oxygen, how much heat (in kilojoules) is liberated?

Explanation / Answer

wt of methane = 0.915* 15.5 = 14.18g

wt of ethane = .085* 15.5= 1.3175 g..

moles of methane = wt of methane / molecular wt = 14.18/16 =.886

moles of ethane = wt of ethane / molecular wt of ethane = 1.3175/ 30 = .0439

hence total moles in sample = 0.886+.0439 = 0.929 moles...

2) p= nrt /v = .929* 8.313*(273+19.95)/(11.7*0.001) = 193366.4 pa= 1.93 bar = 1.93*.986= 1.9 atm

3)partial press of methane = 1.93* .886/.929= 1.84 bar = 1.84* .986 = 1.81 atm

nd of ethane = 1.93- 1.84= 0.089 bar = .089* .986= .087 atm

4)heat liberated = mass * calorific value = 15.5* 10^-3 * 50,000 = 775 kj..

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