Question
Transcribed Text
1. 5 kg of benzene are mixed with 5kg of hexane.
(a) Determine the mole fractions for the mixture.
(b) Determine the average molecular weight of the mixture.
(c) In a steady-state mixing unit, 15 mol/min of the mixture described above is mixed
with 3 mol/min of pure benzene. What are the mol fractions in the existing system?
2. A seperation of benzene, toluene, and decane follows the schematic below
The feed, F = 100 mol/hr, is 30% benzene, 43% toluene, and 27% decane. Stream G conatins no
benzene, 96% of the decane in feed F, and is 99 mol% decane. Stream E contains 95% of the toluene
that enters in feed F. Stream D contains 28 mol/hr benzene, and no decane. Determine the total molar
flowrates of streams D, E, G, and B.
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Q 15 kg of benzene are mixed with 5 kg of hexane.
Q 1(a) Determine the mole fractions for the mixture.
A 1(a) The molecular weight of benzene is 78.11 kg/kmole. The molecular weight of hexane is 86.18 kg/kmol. Thus we have: 5kg/(78.11 kg/kmole)=0.064 kmole=64 mole benzene and 5kg/(86.18 kg/kmole)=0.058 kmole =58 mole hexane. Their ratio in the mixture is: 64:58=52.5%:47.5% benzene:hexane....
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