Chemistry

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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.

Find:

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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