To calculate the yield of benzocaine, we first need to determine the limiting reagent in the reaction. We will do this by calculating the moles of each reactant and comparing their mole ratios.The balanced chemical equation for the formation of benzocaine C9H11NO2 from p-aminobenzoic acid C7H7NO2 and ethanol C2H5OH is:C7H7NO2 + C2H5OH C9H11NO2 + H2OThe molecular weights of the reactants and product are as follows:- p-aminobenzoic acid C7H7NO2 : 137.14 g/mol- ethanol C2H5OH : 46.07 g/mol- benzocaine C9H11NO2 : 165.19 g/molNow, we will calculate the moles of each reactant:- Moles of p-aminobenzoic acid: 5 g / 137.14 g/mol = 0.0364 mol- Moles of ethanol: Since the density of ethanol is approximately 0.789 g/mL, the mass of 10 mL of ethanol is 10 mL 0.789 g/mL = 7.89 g. Therefore, the moles of ethanol are 7.89 g / 46.07 g/mol = 0.1712 mol.Now, we will compare the mole ratios of the reactants to determine the limiting reagent. The stoichiometry of the reaction is 1:1, so we will divide the moles of each reactant by their respective stoichiometric coefficients:- p-aminobenzoic acid: 0.0364 mol / 1 = 0.0364- ethanol: 0.1712 mol / 1 = 0.1712Since 0.0364 is smaller than 0.1712, p-aminobenzoic acid is the limiting reagent.Now, we can calculate the theoretical yield of benzocaine using the moles of the limiting reagent and the stoichiometry of the reaction:Theoretical yield of benzocaine = moles of p-aminobenzoic acid 1 mol benzocaine / 1 mol p-aminobenzoic acid 165.19 g/mol benzocaineTheoretical yield of benzocaine = 0.0364 mol 165.19 g/mol = 6.01 gAssuming 100% conversion efficiency and pure product formation, the yield of benzocaine that can be obtained is 6.01 grams.