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semibatch reaction:

C6H6(l) +Cl2(g) ==> C6H5Cl(l) + HCl(g)

C6H5(l) + Cl2(g) ==> C6H4Cl2(l) + HCl(g)

C6H4Cl2(l) + Cl2(g) ==> C6H3Cl3(l) + HCl(g)

\reactor contained 11.0 mole of benzene and a product mixture in which the amt of mono-chlorobenzene was 3x the amt of dichlorobenzene, and the amount of the latter was 2x the amount of tri-chlorobenzene. The total amount of chlorine fed during the operation was 41.0 mol Cl2.


a) The fractional conversion of benzene.

b) The amt of each liquid component besides benzene in the reactor

c) the amount of HCl produced

d) the fractional conversion of chlorine

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