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Toluene reacts with H2 to form benzene (B), but a side reaction occurs in which a by-product dipheny
90daytona
90daytona
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5 years ago
Toluene reacts with H2 to form benzene (B), but a side reaction occurs in which a by-product dipheny
Toluene reacts with H2 to form benzene (B), but a side reaction occurs in which a by-product diphenyl (D) is formed:
C7H8 + H2 ------------------> C6H6 + CH4 (a)
Toluene hydrogen benzene methane
2C7H8 + H2 ------------------> C12H10 + 2 CH4 (b)
The process is shown in the figure below. Hydrogen is added to the gas recycle stream to make the ratio of H2 to Ch4 1 to 1 before the gas enters the mixer. The ratio of H2 to toluene entering the reactor at G is 4H2 to 1 toluene. The conversion of toluene to benzene on one pass through the reactor is 80%, and the conversion of toluene to the by-product diphenyl is 8% per pass. Calculate the moles of RG and moles RL per hour.
Data: Compound: H2 CH4 C2H6 C7H8 C12H10
2 16 78 92 154
CHE 200 -Chemical Engineering Fundamentals Assignment question
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cloveb
wrote...
#1
Valued Member
5 years ago
Basis: 100 mol toluene (T) in stream G
consider:
a) system: Reactor + Separator
input - output + generation - consumption = 0
for toluene: 100 - RL + 0 - 100(0.8 + 0.08) = 0
for H2: 100(4) - nwH2 + 0 - 100(0.8 + 0.08) = 0
for CH4: 100(4) - nwCH4 + 100(0.8 + 0.08) = 0
RL = 12 mole
nwH2 = 316 mole
nCH4 = 488 mole
Total = 804 mole
b) system: Mixer + Make up point
for toluene: F+RL=100
for H2= M+RG(316/804)=100(4)
for CH4= RG(488/804)=100(4)
RG=659 mol
F=88 mol
M= 141 mol
moles of RG per hour?
(3450 lb/ hr) * (lbmole/ 92 lb) = 37.5 lbmol/ hr toluene in F
(100 lbmol in G/ 88 lbmol in F) * (37.5 lbmol F/ 1 hr)* (659 lbmol RG/100 lbmol G) = 280.8 lbmol RG/hr
moles of RL per hour?
(100 lbmol in G/ 88 lbmol in F) * (37.5 lbmol F/ 1 hr) * (12 lbmol RL/ 100 lbmol in G) = 5.11 lbmol RL/hr
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90daytona
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4 years ago
Thank you for your help
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marukunate
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#3
11 months ago
Thanks
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Anonymous
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2 months ago
Thank you
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Mohammad M.
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2 months ago
Thx
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