Question

a) What is the theoretical yield of ethyl chloride in the reaction of 19.9 g of ethylene with 50 g of hydrogen chloride? (For ethylene, MW=28.0amu; for hydrogen chloride, MW=36.5amu; for ethyl chloride, MW=64.5amu.)

H2C=CH2+HCl→CH3CH2Cl

b) What is the percent yield if 24.0 g of ethyl chloride is actually formed?

Answer 1

number of moles of ethylene = 19.9g / 28.0 g.mol^-1 = 0.711 mole

number of moles of hydrogen chloride = 50g/ 36.5 g.mol^-1 = 1.37 mole

from the balanced equation we can say that

1 mole of ethylene requires 1 mole of HCl so

0.711 mole of ethylene will require 0.711 mole of HCl

ethylene is the limiting reactant

1 mole of ethylene produces 1 mole of ethylchloride

so 0.711 mole of ethylene will produce 0.711 mole of ethylchloride

1 mole of ethylchloride = 64.5g

0.711 mole of ethylchloride = 45.9 g

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

percent yield = (24.0g / 45.9)*100

percent yield = 52.3 %