answer:To calculate the percentage yield, we need to compare the actual yield (4.2 g of ethyl acetate) to the theoretical yield (the maximum possible yield based on stoichiometry). From the balanced equation: ethanol + ethanoic acid → ethyl acetate The stoichiometric ratio between ethanol, ethanoic acid, and ethyl acetate is 1:1:1. First, we need to determine the limiting reagent. This is done by comparing the amount of reactants to the stoichiometric ratio. For ethanol: Molar mass of ethanol (C2H5OH) = 46.07 g/mol Number of moles of ethanol = 6.7 g / 46.07 g/mol ≈ 0.145 mol For ethanoic acid: Molar mass of ethanoic acid (CH3COOH) = 60.05 g/mol Number of moles of ethanoic acid = 5.5 g / 60.05 g/mol ≈ 0.092 mol Since the stoichiometric ratio calls for an equal number of moles of ethanol and ethanoic acid (1:1), ethanoic acid is the limiting reagent in this case. Now, let's calculate the theoretical yield of ethyl acetate based on the limiting reagent: Given that the molar mass of ethyl acetate (CH3COOC2H5) is 88.11 g/mol, we can find the theoretical yield: Theoretical yield = Number of moles of ethanoic acid × molar mass of ethyl acetate = 0.092 mol × 88.11 g/mol ≈ 8.094 g Finally, we can calculate the percentage yield by dividing the actual yield (4.2 g) by the theoretical yield (8.094 g) and multiplying by 100: Percentage yield = (4.2 g / 8.094 g) × 100 ≈ 51.9% Therefore, the percentage yield for the given reaction is approximately 51.9%.

answer:To calculate the percentage yield, we need to compare the actual yield (13.6 g of acetic anhydride) to the theoretical yield (the maximum possible yield based on stoichiometry). From the balanced equation: 2 CH3COOH → H3C-CO-O-CO-CH3 + H2O The stoichiometric ratio between ethanoic acid and acetic anhydride is 2:1. First, we need to determine the limiting reagent. This is done by comparing the amount of reactants to the stoichiometric ratio. For ethanoic acid: Molar mass of ethanoic acid (CH3COOH) = 60.05 g/mol Number of moles of ethanoic acid = 19.7 g / 60.05 g/mol ≈ 0.328 mol For acetic anhydride: Molar mass of acetic anhydride (H3C-CO-O-CO-CH3) = 102.09 g/mol Number of moles of acetic anhydride = 13.6 g / 102.09 g/mol ≈ 0.133 mol Since the stoichiometric ratio calls for double the number of moles of ethanoic acid compared to acetic anhydride (2:1), acetic anhydride is the limiting reagent in this case. Now, let's calculate the theoretical yield of acetic anhydride based on the limiting reagent: Given that the molar mass of acetic anhydride (H3C-CO-O-CO-CH3) is 102.09 g/mol, we can find the theoretical yield: Theoretical yield = Number of moles of acetic anhydride × molar mass of acetic anhydride = 0.133 mol × 102.09 g/mol ≈ 13.59 g Finally, we can calculate the percentage yield by dividing the actual yield (13.6 g) by the theoretical yield (13.59 g) and multiplying by 100: Percentage yield = (13.6 g / 13.59 g) × 100 ≈ 99.93% Therefore, the percentage yield for the given reaction is approximately 99.93%.

answer:The correct solution to the quadratic equation x^2 + 2x - 11 = 0 can be found by factoring, completing the square, or using the quadratic formula. Here are four possible solutions: 1) x = -4, 3 2) x = -6, 5 3) x = -11, 1 4) x = -3, 4 You need to select just one answer.

answer:The quadratic equation x^2 + 2x - 11 = 0 can be solved using the quadratic formula, which is given by: x = (-b ± √(b^2 - 4ac)) / (2a) For this equation, a = 1, b = 2, and c = -11. Substituting these values into the quadratic formula gives: x = (-(2) ± √((2)^2 - 4(1)(-11))) / (2(1)) Simplifying further: x = (-2 ± √(4 + 44)) / 2 x = (-2 ± √48) / 2 x = (-2 ± 4√3) / 2 x = -1 ± 2√3 Therefore, the correct solutions to the given quadratic equation in quadratic form are: x = -1 + 2√3 x = -1 - 2√3