A multiple unit process involves the production of toluene (C6H5CH3) from benzen

Question

A multiple unit process involves the production of toluene (C6H5CH3) from benzene using a Friedel-Crafts reaction technique. Benzene (C6H6) is reacted with methyl chloride (CH3Cl) in the presence of aluminum chloride (AlCl3) catalyst to produce toluene and hydrochloric acid (HCl). Benzene and methyl chloride are fed to the reactor along with 66.7 kg AlCl3. The limiting reactant is completely used up in this reaction. The products leaving the reactor are fed into a separator where the AlCl3 and HCl are removed. All of the AlCl3 is recycled back to the feed stream into the reactor. All of the HCl is removed and prepared for proper waste disposal. The remaining materials leaving the separator are fed into a vaporizer where the toluene and benzene are separated. The stream entering the vaporizer contains only benzene and toluene. The vaporizer is set at a temperature of 95°C and the pressure is fixed at 1 atm. 90% (molar basis) of the liquid that enters the vaporizer ends up in the liquid (bottom) stream. All of the vapor is collected in the top stream and has 57.9 wt% benzene.

Molar masses: C6H6 = 78.11; CH3Cl = 50.49; C6H5CH3 = 92.14; HCl = 36.46

a) If 100 kg of toluene is desired in the liquid stream coming out of the vaporizer, and the toluene has a mole fraction of 0.60 in this stream, what is the composition of the stream entering the vaporizer?

b) Using the basis of 100 kg of toluene in the liquid stream coming out of the vaporizer, how much benzene (mass in kg) must be fed to the reactor? What is the fractional excess of benzene?

c) Determine the total number of moles and composition of the stream leaving the reactor.

d) Will a buildup of the catalyst occur in the reactor?

Explanation / Answer

(a). Mass of toluene = 100 kg

Mole fraction of toluene = 0.60

Moles of toluene = 1.085 kmole

Let moles of benzene in out stream = nB

Mole fraction of toluene = 1.085 / 1.085 + nB = 0.60

1.085 = 0.651 + 0.60nB

0.434 = 0.60nB

nB = 0.723 kmol

Mass of benzene in outstream = 723.3 * 78.11

= 56.49 kg

Hence composition of the stream entering the vaporizer is -

Benzene = 0.7233 kmol =  56.49 kg

Toluene = 1.085 kmole = 100 kg

(b). C6H6 + CH3Cl ---> C5H5CH3 + HCl

Moles of toluene produced = 1.085 kmole

As 1 mole benzene produce 1 mole toluene. So,

Moles of benzene fed to the reactor = 1.085 kmole

Mass of benzene fed to the reactor = 1.085 * 78.11

= 84.75 kg

Fractional excess of benzene = 56.49 / 84.75

= 0.66

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