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    So a reaction between benzene and chlorine occurs forming dichlorbenzene and monochlorobenzene. The question states:

    Assume that the molar ratio of the feed of chlorine to that of benzene is 0.9:1 at the inlet to the reactor. For a feed of 100 kmol/h of benzene to the inlet of the reactor, calculate the molar flow of all the species leaving the reactor.

    The overall conversion of benzene is 55.3%, of which 73.6% reacts to form monochlorobenzene and the remainder form dichlorobenzene.
    So first I got the reaction equations:

    C6H6 + Cl2 --> C6H6Cl + HCl
    C6H6 + 2Cl2 --> C6H4Cl2 + 2HCl

    Then using the mole ratios I got:

    • Moles of benzene in: 47.4 kmol/h
    • Moles of chlorine in: 52.6 kmol/h

    Then, if I remember correctly, conversion is:

    X = consumed/fed

    so the amount of benzene consumed is: 25.122 kmol/h and hence 47.4-25.122=22.3 kmol/h leaves the reactor.

    Here's where I get a bit stuck. If 55.3% of benzene is consumed, how much Cl2 is consumed? Do I have to take into consideration the the amount of moles (i.e. the coefficient of the Cl2 is the reaction equations)?

    And also, would the amount of monochlorobenzene out simply be 73.6% of the benzene that is reacted and the rest be dichlorobenzene?


    1 mol of Cl2 is consumed per every monochlorobenzene.
    1 further mol of Cl2 is consumed per every disubstitution.

    I would;

    1. Work out the total number of moles of benzene consumed
    2. =n1(Cl2)
    3. n2(Cl2)= moles of dichlorobenzene
    4. ntotal(Cl2)=n1(Cl2)+n2(Cl2)
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Updated: May 10, 2015


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