A mixture is 11 mol% ethanol, 79 mol% ethyl acetate (C4H8O2) and 10 mol% acetic

Question

A mixture is 11 mol% ethanol, 79 mol% ethyl acetate (C4H8O2) and 10 mol% acetic acid (C2H4O2).

(a) Calculate the mass fraction of ethanol in this three-component mixture.

(b) What is the average molecular weight of the three-component mixture?

(c) What would be the mass (kg) of the three-component mixture such that the mixture contains 24 kmol of ethyl acetate?

(d) A stream of the three-component mixture enters a mixing drum at 78 kg/min. For a basis of 78 kg/min, what is the inlet molar flowrate (kmol/min) for the mixture?

(e) Also entering the mixing drum described in part (d) is a stream of pure ethanol. The stream exiting the mixing drum is 15 mol% ethanol. Using the basis from part (d), what is the flow rate of pure ethanol into the mixer (kmol/min)?

Explanation / Answer

Basis :1 mole of mixture of ethanol, ethylacetate and acetic acid

Moles : ethanol =0.11, ethyl acetate =0.79 and acetic acid =0.1

Molecular weghts : ethanol (C2H5OH)= 46, ethyl acetate (C4H8O2)= 4*12+8+32= 88 and C2H4O2= 2*12+4+32= 60

Mole= mass /molecular weight

Mass =mole* Molecular weight

Masses : Ethanol =0.11*46 = 5.06 gms Ethyl acetate= 88*0.79 gms=69.52 gms and acetic acid = 0.1*60= 6 gms

Total mass= 5.06+69.52+6=80.58 gms

Molecular weight= 80.58 g/mole

24 kmol of ethyl acetate contains 24*5.06/0.79 kg of ethanol=153.712kg and 24*6/0.79=182.3 kg of acetic acid

78 kg/min correspond to 78/80.58 kgmole/min =0.968 kgmole/min

Let moles of ethanol added = F

Total moles of ethanol = F+0.10648 mol/min

Moles of mixture exiting= 0.968+F

Exit mixture contains 15 mole% ethanol

F+0.10648= (0.968+F)*0.15

F +0.10648= 0.1452 +0.1F

0.9F= 0.1452-0.10648=0.03872

F= 0.03872/0.9 =0.043 mole/min

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