in the the lab experiment of Friedel-Crafts Alkylation: 0.4 ml of benzene was mixed with 0.8...
in the the lab experiment of Friedel-Crafts Alkylation: 0.4 ml of benzene was mixed with 0.8 ml of 2-chloro 2 methyl propane (the ratio is 1 to 2) to produce 1-4 dibutyl benzene. how to calculate percent yeild and what is the limiting reagent?
Homework Answers
Determination of limiting reagent :
Number of moles of benzene =
weight(grams)/molecular weight
Weight of benzene = volume* density
= 0.4 ml * 0.876
= 0.3504 grams
Moles of benzene = 0.3504/78
= 0.00449 moles (or) 4.49 m.moles.
Number of moles of 2-chloro-2-methyl propane = weight/molecular weight
Weight of 2-chloro-2-methylpropane = density * volume
= 0.851*0.8
= 0.6808 grams.
Moles of 2-chloro-2-methylpropane = 0.6808/92.57
= 0.007354 moles (or) 7.3544 m.moles.
Among the two reactants benzene has less number of moles.thus benzene is the limiting reagent.
Calculation of theoretical yield:
78 grams of benzene produces 190.32 grams of 1,4-di butyl benzene.
0.3504 grams of benzene produces ? grams of 1,4- diDbutyl benzene
= 0.3504*190.32/78
= 0.854 grams.
Theoretical yield (100% yield) = 0.854 ki.
% of yield = experimental weight (grams)*100/theoretical yield
But experimental weight is not given in the data.
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