Question

7.4 Theoretical and Percent Yield 10. A chemist carries out the following reaction: CH + Br2 → C.H.Br + HBr (2 points each) a. If he reacts 14.2 grams of CoHs with an excess of Brz, what is the theoretical yield of CeHsBr? 14:29 CGHE X Imol CGHE & Imol 78. 112g C6H6 - b. If he collected 16.3 grams of CHsBr after finishing his experiment, what is his percent yield?

Answer 1

10)

a)

C6H6 + Br2 ---------------- C6H5Br + HBr

1 mole                            1 mole

mass of C6H6 = 14.2 grams

molar mass of C6H6 = 78.11 gram/mole

number of moles of C6H6 = mass / molar mass = 14.2 / 78.11 = 0.182 moles

number of moles of C6H6 = 0.182 moles

according to equation

1 mole of C6H6 = 1 mole of C6H5Br

0.182 moles of C6H6 = ?

                                    = 0.182 x 1/1 = 0.182 moles of C6H5Br

number of moles of C6H5Br formed = 0.182 moles

molar mass of C6H5Br = 157 gram/mole

mass of one mole of C6H5Br = 157 gram

mass of 0.182 moles of C6H5Br = 0.182 x 157 = 28.574 grams

Theoritical yield of C6H5Br = 28.57 grams

B)

Actual yield of C6H5Br = 16.3 grams

Theoritical yield of C6H5Br = 28.57 grams

Percent yield = Actual yield x100 / Theoritical yield

Percent yield = 16.3 x100 / 28.57 = 57.05%

Percent yield of C6H5Br = 57.05%