2 0268 in 100 to compute the fractional yield of

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Unformatted text preview: d of monochlorobenzene we need to know how much monochlorobenzene would have formed if all of the limiting reactant (Cl2) had reacted and no by-products had been formed. Since there are 28.7 mol of Cl2 and the stoichiometry for the formation of monochlorobenzene is 1:1 with Cl2 28.7 mol of C6H5Cl should have been formed if all of the Cl2 reacted to give monochlorobenzene. The fractional yield is thus yield mol formed 25 0.871 mol formed without by products 28.7 The liquid feed is the benzene. Thus the mass of the benzene is g mbenz 100 mol 78 7800 g mol To get the mass of the gas phase first compute the mass of chlorine fed g mchlorine 28.7 mol 70.9 2035 g mol Since Cl2 is 98 wt% of the gas the gas phase mass must be 2035/0.98 = 2076 g. So the requested mass ratio is m gas mliquid 2076 0.266 7800 (c) The benzene is fed in excess to consume as much of the Cl2 as possible. Keeping the Cl2 level low reduces the rate of formation of the by-products. Similarly, a low conversion of C6H6 keeps the concentration of C6H5Cl low,...
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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.

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