Unformatted text preview: of the species, including the
impurities in the technical grade chlorine.
(b) Take 100 mole of the product gas as a basis. Then use the mole fractions determined
above to calculate the number of moles of reactant (C6H6 and Cl2) need to produce that
number of moles of product. This would give the molar composition of the reactant
C6H3Cl3 Mole Frac.
0.242 Moles C6H6
0.242 Moles Cl2
0.726 So the reactant stream would have had to contain 73.2+25.0+1.49+0.242 = 100 mol C6H6
and 25.0+2.98+0.726 = 28.7 mol Cl2. If the basis is the formation of monochlorobenzene
we need one mol of Cl2 per mol of benzene. Since there were 28.7 mol Cl2 the amount of
benzene needed (stoichiometrically) is also 28.7 mol. So the percentage of benzene in
excess 100 28.7
28.7 The fractional conversion of C6H6 is
conversion in out 100 73.2 0.268
100 To compute the fractional yiel...
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This note was uploaded on 03/01/2013 for the course CHEME 2171 taught by Professor Benton during the Spring '12 term at LSU.
- Spring '12