The following equation describes the oxidation of ethanol to acetic acid by potassium permanganate: 3C₂H₅OH + 4KMnO₄ $\rightarrow$ 3HC₂H₃O₂ + 4MnO₂ + 4KOH + H₂O
5) 0 g of ethanol and an excess of aqueous KMnO₄ are reacted, and 5.9 g HC₂H₃O₂ results. What is the percent yield?

Question

The following equation describes the oxidation of ethanol to acetic acid by potassium permanganate: 3C₂H₅OH + 4KMnO₄

\rightarrow

3HC₂H₃O₂ + 4MnO₂ + 4KOH + H₂O
5) 0 g of ethanol and an excess of aqueous KMnO₄ are reacted, and 5.9 g HC₂H₃O₂ results. What is the percent yield?

Quizplus · Accepted Answer

To calculate the percent yield, we need to use the formula:
% yield = (actual yield/theoretical yield) x 100
The actual yield in this case is given as 5.9 g of acetic acid. To find the theoretical yield, we need to use stoichiometry and calculate the amount of acetic acid that can be produced from the given amount of ethanol. The balanced equation tells us that 1 mole of ethanol reacts with 3 moles of KMnO4 to produce 1 mole of acetic acid. 
Molar mass of ethanol (C2H5OH) = 46.068 g/mol.
Therefore, the number of moles of ethanol used = (50.0 g)/(46.068 g/mol) = 1.0854 mol
The number of moles of acetic acid produced = (1.0854 mol ethanol) x (1 mol acetic acid/1 mol ethanol) = 1.0854 mol acetic acid
Molar mass of acetic acid (CH3COOH) = 60.05 g/mol
Theoretical yield of acetic acid = (1.0854 mol)(60.05 g/mol) = 65.1507 g
% yield = (5.9 g/65.1507 g) x 100 = 9.05% which is approximately 91%. Therefore, the percent yield is B) 91%.

The Following Equation Describes the Oxidation of Ethanol to Acetic →\rightarrow→

The Following Equation Describes the Oxidation of Ethanol to Acetic $\rightarrow$