What is the percent yield of the reaction if you obtained 1.25 g of p-bromoacetanilide from the reaction of 1.00 g of acetanilide with 1.80 g of Na Br and excess NaOCl solution?

The chemical reaction is written as:

2H + 2NaBr + 2C8H9NO + NaClO - - > NaCl + 2C8H9NOBr + H2O

Percent yield = actual yield / theoretical yield x 100

theoretical yield = 1.00 g acetanilide (1 mol / 135.17 g) (2 mol

p-bromoacetanilide / 2 acetanilide) (214.062 g / mol) = 1.58 g

Percent yield = actual yield / theoretical yield x 100

Percent yield = 1.25 / 1.58 x 100

Percent yield = 79.11%