To calculate the maximum yield of p-nitroaniline, we need to use stoichiometry. First, we need to find the molar mass of aniline C6H7N and p-nitroaniline C6H6N2O2 .Molar mass of aniline:C: 12.01 g/mol x 6 = 72.06 g/molH: 1.01 g/mol x 7 = 7.07 g/molN: 14.01 g/mol x 1 = 14.01 g/molTotal molar mass of aniline = 72.06 + 7.07 + 14.01 = 93.14 g/molMolar mass of p-nitroaniline:C: 12.01 g/mol x 6 = 72.06 g/molH: 1.01 g/mol x 6 = 6.06 g/molN: 14.01 g/mol x 2 = 28.02 g/molO: 16.00 g/mol x 2 = 32.00 g/molTotal molar mass of p-nitroaniline = 72.06 + 6.06 + 28.02 + 32.00 = 138.14 g/molNow, we can use stoichiometry to find the maximum yield of p-nitroaniline. The balanced chemical equation for the reaction is:C6H7N + HNO3 C6H6N2O2 + H2OFrom the balanced equation, we can see that 1 mole of aniline reacts with 1 mole of nitric acid to produce 1 mole of p-nitroaniline.First, we need to find the moles of aniline:50 g aniline 1 mol aniline / 93.14 g aniline = 0.536 moles of anilineSince the reaction goes to completion and there is an excess of nitric acid, all of the aniline will be converted to p-nitroaniline. Therefore, we can find the mass of p-nitroaniline produced:0.536 moles of aniline 1 mol p-nitroaniline / 1 mol aniline 138.14 g p-nitroaniline / 1 mol p-nitroaniline = 74.08 g of p-nitroanilineSo, the maximum yield of p-nitroaniline that can be obtained when 50g of aniline is reacted with an excess of nitric acid is 74.08 g.