1. Engineering
  2. Chemical Engineering
  3. 3 natural gas consisting of 94000 moles methane 3000 moles...

Question: 3 natural gas consisting of 94000 moles methane 3000 moles...

Question details

3. Natural gas consisting of 94,000 moles methane, 3,000 moles ethane, 2,000 moles propane and 1,000 moles butane is bubbled through 10,000 moles of initially pure hexane at 225 K and 25.0 bar total pressure. The gas leaving the scrubber contains 88570 moles methane, 1560 moles ethane, 248 moles propane and 19 moles butane To see if Raoults law is obeyed, compare the partial pressure of butane(4) in the scrubbed gas (p4 y4p) with the pressure of butane calculated from Raoults law (p4 xp). Data: The a) 17] vapor pressure of pure liquid butane at 225 K is p4* 0.106 bar. The vapor pressure of liquid hexane is negligible at this temperature (p0, y6 0). b) c) Calculate the percentage of the total ethane, propane and butane removed from the natural gas. Explain why is the percentage of butane removed is much higher than the percentage of ethane removed. (Data: at 225 K, p:-5.88 bar for pure liquid ethane)
Solution by an expert tutor
Blurred Solution
This question has been solved
Subscribe to see this solution