The reaction of aniline with chlorine gas to prepare 2-chloro-4-nitroaniline involves two steps. First, aniline reacts with a nitrating agent, such as a mixture of concentrated nitric acid HNO3 and concentrated sulfuric acid H2SO4 , to form 4-nitroaniline. Then, 4-nitroaniline reacts with chlorine gas to form 2-chloro-4-nitroaniline.Step 1: Nitration of anilineC6H7N + HNO3 C6H6N2O2 + H2OStep 2: Chlorination of 4-nitroanilineC6H6N2O2 + Cl2 C6H5ClN2O2 + HClOverall reaction:C6H7N + HNO3 + Cl2 C6H5ClN2O2 + H2O + HClNow, let's calculate the percent yield:1. Calculate the theoretical yield:Molar mass of aniline C6H7N = 93.13 g/molMolar mass of 2-chloro-4-nitroaniline C6H5ClN2O2 = 172.57 g/mol10 g of aniline * 1 mol aniline / 93.13 g * 1 mol 2-chloro-4-nitroaniline / 1 mol aniline * 172.57 g / 1 mol 2-chloro-4-nitroaniline = 18.53 g theoretical yield 2. Calculate the percent yield:Percent yield = actual yield / theoretical yield * 100Percent yield = 7 g / 18.53 g * 100 = 37.78%The balanced chemical equation for the reaction is:C6H7N + HNO3 + Cl2 C6H5ClN2O2 + H2O + HClThe percent yield of the reaction is 37.78%.