Question

the following equation describes the oxidation of ethanol to acetic acid by potassium permanganate:
$3\ce{c2h5oh} + 4\ce{kmno4} \longrightarrow 3\ce{hc2h3o2} + 4\ce{mno2} + 4\ce{koh} + \ce{h2o}$

$5.00\\ \text{g}$ of ethanol and an excess of aqueous $\ce{kmno4}$ are reacted, and $4.24\\ \text{g}$ of $\ce{hc2h3o2}$ result. what is the percent yield?

• 1. $34.9\\%$
• 2. $21.7\\%$
• 3. $100\\%$
• 4. $4.24\\ \text{g}\\ \ce{hc2h3o2}$ is impossible since it represents more than $100\\%$ yield
• 5. $65.1\\%$

Explanation:

Step1: Calculate moles of ethanol

Molar mass of \( C_2H_5OH \) is \( 46.07 \, g/mol \). Moles of ethanol = \( \frac{5.00 \, g}{46.07 \, g/mol} \approx 0.1085 \, mol \).

Step2: Determine moles of acetic acid (theoretical)

From the balanced equation, 3 moles of \( C_2H_5OH \) produce 3 moles of \( HC_2H_3O_2 \). So moles of \( HC_2H_3O_2 \) (theoretical) = moles of ethanol = \( 0.1085 \, mol \).

Step3: Calculate theoretical mass of acetic acid

Molar mass of \( HC_2H_3O_2 \) is \( 60.05 \, g/mol \). Theoretical mass = \( 0.1085 \, mol \times 60.05 \, g/mol \approx 6.51 \, g \).

Step4: Calculate percent yield

Percent yield = \( \frac{actual \, yield}{theoretical \, yield} \times 100 = \frac{4.24 \, g}{6.51 \, g} \times 100 \approx 65.1\% \). Wait, no, wait: Wait, 3 moles of ethanol give 3 moles of acetic acid, so mole ratio 1:1. Wait, my mistake earlier: Let's recalculate theoretical yield. Moles of ethanol: \( 5.00g / 46.07g/mol = 0.1085mol \). Theoretical moles of acetic acid: 0.1085mol. Theoretical mass: 0.1085mol 60.05g/mol = 6.51g. Actual yield is 4.24g. Wait, no, wait, 4.24g actual, 6.51g theoretical? Wait, no, that can't be. Wait, no, I think I messed up the mole ratio. Wait the balanced equation: 3 \( C_2H_5OH \) produces 3 \( HC_2H_3O_2 \), so mole ratio 1:1. So theoretical mass: moles of ethanol molar mass of acetic acid. Wait, 5.00g ethanol is 5/46.07 = 0.1085mol. So theoretical acetic acid is 0.1085mol 60.05g/mol = 6.51g. Then percent yield is (4.24 / 6.51)100 ≈ 65.1%? Wait, but the options have 65.1% as option 5. Wait, but let's check again. Wait, maybe I made a mistake in mole ratio. Wait the equation: 3 \( C_2H_5OH \) + 4 \( KMnO_4 \) → 3 \( HC_2H_3O_2 \) + 4 \( MnO_2 \) + 4 \( KOH \) + \( H_2O \). So 3 moles ethanol → 3 moles acetic acid, so 1:1. So yes, theoretical yield is 6.51g, actual is 4.24g. Wait, no, wait 4.24g actual, 6.51g theoretical? Then (4.24/6.51)100 = 65.1%? Wait, but the options have 65.1% as option 5. Wait, but let's recalculate: 5.00g ethanol: moles = 5/46.07 = 0.1085mol. Theoretical acetic acid: 0.1085mol 60.05g/mol = 6.51g. Actual yield 4.24g. Percent yield = (4.24 / 6.51)100 = 65.1%. So the answer is 65.1%, option 5. Wait, but wait, maybe I messed up the molar mass. Wait ethanol molar mass: C=12.01, H=1.008, O=16.00. So \( C_2H_5OH \): 212.01 + 61.008 + 16.00 = 24.02 + 6.048 + 16.00 = 46.068 ≈ 46.07g/mol. Acetic acid: \( HC_2H_3O_2 \) is \( CH_3COOH \), molar mass: 12.012 + 41.008 + 216.00 + 1.008 = 24.02 + 4.032 + 32.00 + 1.008 = 61.06g/mol? Wait, no! Wait \( HC_2H_3O_2 \) is acetic acid, formula \( C_2H_4O_2 \), molar mass: 212.01 + 41.008 + 216.00 = 24.02 + 4.032 + 32.00 = 60.052 ≈ 60.05g/mol. Correct. So theoretical yield: 0.1085mol 60.05g/mol = 6.51g. Actual yield 4.24g. Percent yield: (4.24 / 6.51)*100 = 65.1%. So the answer is 65.1%, option 5.

Answer:

1. 65.1%