Question

In part D of this experiment, you are to use 11.2 g of 4-bromo-2-chloroacetanilide (MM = 248.5 g/mol). What is the theoretical yield of 4-bromo-2-chloroaniline (MM = 206.47 g/mol)?

a. 9.3 g

b. 11.2

c. 10.4

d. 9.8 g

In part E of this experiment, you are to use 2.5 g of 4-bromo-2-chloroaniline (MM = 206.47 g/mol) and 2.5 g of ICl (MM = 162.35 g/mol). What is the theoretical yield of 4-bromo-2-chloro-5-iodoaniline (MM = 332.36 g/mol)?

a. 4.0g

b. 5.0 g

c. 3.5 g

d. 6.0 g

Answer 1

Part d

1 mol 4-bromo-2-chloroacetanilide produce 1 mol of 4-bromo-2-chloroaniline

248.5 g  4-bromo-2-chloroacetanilide produces = 206.47 g of 4-bromo-2-chloroaniline

11.2 g of 4-bromo-2-chloroacetanilide produces = 206.47*11.2/248.5 = 9.3 g of 4-bromo-2-chloroaniline

Option A is the correct answer

Part E

206.47 g of 4-bromo-2-chloroaniline require = 162.35 g of ICl

2.5 g of 4-bromo-2-chloroaniline require = 162.35*2.5/206.47 = 1.96 g of ICl

ICl is the excess reagent and 4-bromo-2-chloroaniline is limiting reagent

theoretical yield of 4-bromo-2-chloro-5-iodoaniline

= 332.36*2.5/206.47 = 4 g

Option A is the correct answer