<Ol> <Li> The molar amount of the reactants is calculated from the weights (acetic acid: 120 g ÷ 60 g / mol = 2.0 mol; ethanol: 230 g ÷ 46 g / mol = 5.0 mol). </Li> <Li> Ethanol is used in a 2.5-fold excess (5.0 mol ÷ 2.0 mol). </Li> <Li> The theoretical molar yield is 2.0 mol (the molar amount of the limiting compound, acetic acid). </Li> <Li> The molar yield of the product is calculated from its weight (132 g ÷ 88 g / mol = 1.5 mol). </Li> <Li> The% yield is calculated from the actual molar yield and the theoretical molar yield (1.5 mol ÷ 2.0 mol × 100% = 75%). </Li> </Ol> <Li> The molar amount of the reactants is calculated from the weights (acetic acid: 120 g ÷ 60 g / mol = 2.0 mol; ethanol: 230 g ÷ 46 g / mol = 5.0 mol). </Li> <Li> Ethanol is used in a 2.5-fold excess (5.0 mol ÷ 2.0 mol). </Li> <Li> The theoretical molar yield is 2.0 mol (the molar amount of the limiting compound, acetic acid). </Li>

Why do most chemical reactions never reach 100 completion