The following equation describes the oxidation of ethanol to acetic acid by potassium permanganate: 5.00 g of ethanol and an excess of aqueous KMnO₄ are reacted,and 4.24 g of HC₂H₃O₂ result.What is the percent yield? A)100% B)65.1% C)21.7% D)34.9% E)4.24 g HC₂H₃O₂ is impossible since it represents more than 100% yield.

Question

Quizplus · Accepted Answer

The percent yield is calculated by dividing the actual yield (4.24 g) by the theoretical yield (calculated based on the amount of ethanol used) and multiplying by 100%. The balanced equation tells us that 2 moles of ethanol react with 1 mole of KMnO4 to produce 2 moles of acetic acid. From the molar masses, we can calculate that 5.00 g of ethanol is equivalent to 0.104 mol. Therefore, the theoretical yield of acetic acid is 0.104 mol x 2 mol acetic acid/2 mol ethanol x 60.05 g/mol acetic acid = 6.02 g. The percent yield is (4.24 g / 6.02 g) x 100% = 70.4%, which is closest to choice B (65.1%).

The Following Equation Describes the Oxidation of Ethanol to Acetic