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%

A) the enthralling of formation for pure element in its standard state is always positivehope this : )