Question

If the hydrolysis to the diacid is not complete, how could you separate the desired diacid from the unhydrolyzed anhydride by extraction?

You carry out a reaction starting with 200 mg of cyclopentadiene and 300 mg of maleic anhydride and 375 mg of the cycloadduct anhydride is obtained. Calculate the % yield for the anhydride. Show your calculation.

Answer 1

Solution :-

1) To seprate the unreacted acid from the mixture we need to use the NaHCO3 solution which will precipitate the unreacted acid from the mixture and then we can separate the unreacted acid and the product from the mixture by the filtration method.

2) Solution :-

You carry out a reaction starting with 200 mg of cyclopentadiene and 300 mg of maleic anhydride and 375 mg of the cycloadduct anhydride is obtained. Calculate the % yield for the anhydride. Show your calculation

Balanced reaction equation is as follows

C5H6        +                      C4H2O3     -------------- > C10H8O3

Cyclopentadiene               Maleic unhydride          cycloadduct

MW= 66.1 g per mol        MW =98.06 g per mol       MW= 176.1687 g per mol

Mole ratio of the reaction for both reactants with product is 1 : 1

Lets convert the given masses from mg to gramm

200 mg * 1 g / 1000 mg = 0.200 g cyclopentadiene

300 mg *1 g /1000 mg = 0.300 g maleic unhydride

375 mg *1 g /1000 mg= 0.375 g cycloadduct.

Now using the mole ratio of the each reactant lets calculate the theoretical yield of the product

0.200 g cyclopentadine * 176.1687 g cycloadduct / 66.1 g cyclopentadiene = 0.533 g Cycloadduct

0.300 g maleic unhydride * 176.1687 g cycloadduct / 98.06 g maleic unhydride = 0.538 g cycloadduct

The mass of the cycloadduct produced from the 0.2 g cyclopentadiene is less therefore cyclopenbtadine is the limiting reactant

Therefore the theoretical yield of the product is 0.533 g cycloaduct

Now lets calculate the percent yield

% yield = (actual yield /theoretical yield)*100%

             = (0.375 g / 0.533 g)*100%

            = 70.36 %

Therefore percent yield of the reaction is 70.36 %