Question

What is the percent yield of p-nitroaniline obtained when 10g of aniline is reacted with excess nitric acid to produce p-nitroaniline and 8g of the product is obtained after purification?

Answer 1

To calculate the percent yield of p-nitroaniline, we first need to determine the theoretical yield based on the stoichiometry of the reaction.The balanced chemical equation for the reaction is:C6H7N + HNO3  C6H6N2O2 + H2OAniline  C6H7N  has a molar mass of 93.13 g/mol, and p-nitroaniline  C6H6N2O2  has a molar mass of 138.12 g/mol.First, we need to convert the mass of aniline to moles:10 g aniline   1 mol aniline / 93.13 g aniline  = 0.1073 mol anilineSince the reaction has a 1:1 stoichiometry, the theoretical yield of p-nitroaniline is also 0.1073 mol. Now, we can convert this to grams:0.1073 mol p-nitroaniline   138.12 g p-nitroaniline / 1 mol p-nitroaniline  = 14.82 g p-nitroaniline  theoretical yield Now that we have the theoretical yield, we can calculate the percent yield:Percent yield =  actual yield / theoretical yield   100Percent yield =  8 g p-nitroaniline / 14.82 g p-nitroaniline   100 = 53.98%The percent yield of p-nitroaniline in this reaction is approximately 53.98%.