Question

1. Calculate the percentage yield, if 1g of product is formed when 2.0 g of 3-sulfolene, 2 g of finely powdered maleic anhydride, and 1.0 mL of xylenes are used.

Answer 1

The reaction is given below

Molar mass of 3-Sulfolene = 118 g/mol

molar mass of maleic anhydride = 98 g/mol

molar mass of the product = 152 g/mol

Given mass of 3- sulfolene = 2.00

hence moles = 2.00/ 118 = 0.0169

Given mass of maleic anhydride = 2 g

Moles = 2/98 = 0.0203

From the reaction we can write

1 mole 3 -sulfolene reacts with 1 mole maleic anhydride to produce 1 mole of the product .

as given moles of 3-sulfolene is less than maleic anhydride , then limiting reagent is 3 -sulfolene

so moles of the product formed = moles of 3-sulfolene taken = 0.0169

then theoretical mass of the product = 0.0169 *molar mass of the product = 0.0169*152 = 2.5688 g

Percentage yield = actual yield *100/ theoretical yield

= (1*100)/2.5688 = 38.92 %