Question

114. The following equation describes the oxidation of ethanol to acetic acid by potassium permanganate!
3C,H,OH + 4KMnO4 → 3HC,H,O2 + 4MnO+4KOH + H2O
5.00 g of ethanol (46.07 g/mol) and an excess of aqueous KMnO4 are reacted, and 5.99 g of HC2H302 (60.05
g/mol) result. What is the percent yield?

Answer 1

Answer:

Explanation:

The following equation describes the oxidation of ethanol to aceticacid by potassium permanganate:

3C2H5OH +4KMnO4-->3HC2H3O2 +4MnO2 + 4KOH + H2O

5.0 g of ethanol and an excess of aqueous KMnO4 arereacted, and 5.9 g HC2H3O2 result.What is the percent yield?

a. 100%

b. 91%

c. 67%

d. 30%

e. 5.9 g HC2H3O2 is impossible since it represcents more than a100% yield.

Answer 2

The percent yield of acetic acid is 91.9%

From the question,

We are to determine the percent yield of the acetic acid

To determine the percent yield of the acetic acid, we will calculate the theoretical yield first.

The balanced chemical equation for the reaction is

3C₂H₅OH + 4KMnO₄ → 3HC₂H₃O₂ + 4MnO₂ + 4KOH + H₂O

This means

3 moles of ethanol reacts with 4 moles of potassium permanganate to produce 3 moles of acetic acid

Now, we will determine the number of moles of ethanol that reacted

From the question

Mass of ethanol that reacted = 5.00 g

Molar mass of ethanol = 46.07 g/mol

Using the formula

$Number\ of\ moles= \frac{Mass}{Molar\ mass}$

∴ Number of moles of ethanol = $\frac{5.00}{46.07}$

Number of moles of ethanol that reacted = 0.10853 mole

Now,

From the balanced chemical equation,

3 moles of ethanol reacts with 4 moles of potassium permanganate to produce 3 moles of acetic acid

Then,

0.10853 mole of ethanol reacts with the excess potassium permanganate to produce 0.10853 mole of acetic acid

∴ 0.10853 mole of acetic acid will be produced

Now, for the mass of acetic acid that would be produced

From the formula

Mass = Number of moles × Molar mass

From the question

Molar mass of acetic acid = 60.05 g/mol

∴ Mass of acetic acid that will be produced = 0.10853 × 60.05

Mass of acetic acid that will be produced = 6.5172 g

This is the theoretical yield of acetic acid

Now, for the percent yield

$Percent\ yield = \frac{Actual\ yield}{Theoretical\ yield} \times 100\%$

From the question,

Actual yield = 5.99 g

∴ $Percent\ yield = \frac{5.99}{6.5172} \times 100\%$

$Percent\ yield = \frac{599}{6.5172} \%$

Percent yield = 91.9106 %

Percent yield ≅ 91.9%

Hence, the percent yield of acetic acid is 91.9%

Learn more here: https://brainly.com/question/19655772