Explain step by step lene glycol HOCH CH OH), used as an antifreeze, is produced

Question

Explain step by step lene glycol HOCH CH OH), used as an antifreeze, is produced by reacting ethylene oxide (C,Ho) A side reaction produces an undesirable dimer, DEG with water C,H0+ H20 à HOCH CH OH HOCH CH2OH C2H40 à HOCH2CH20CH CH20 In one preparation, feed of 100 Ibmole/min ethylene oxide and 300 Ibmole/min water are fed to the reactor The conversion of ethylene oxide is found to be 0.92 and DEG leaving the reactor is 69 Ibmol min. Then the afount of ethylene glycol in the stream leaving the reactor is _1bmolmin o 85.1 6.90 O 78.2 92.0

Explanation / Answer

Moles of ethylene oxide C2H4O = 100 lbmol/min

Fractional conversion of ethylene oxide = 0.92

Moles of ethylene glycol produced = moles of ethylene oxide consumed

= 0.92*100 = 92 lbmol /min

From the second stoichiometry of the reaction

1 mol DEG produced from = 1 mole of ethylene glycol consumed

6.9 mol DEG produced from = 6.9 mole of ethylene glycol consumed

Amount of ethylene glycol in the stream leaving the reactor

= ethylene glycol produced from first reaction - ethylene glycol consumed from second reaction

= 92 - 6.9

= 85.1 lbmol/min

Option A is the correct answer

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