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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?

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MissPhiladelphia
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%

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