Question

6.Souring of wine occurs when ethanol is converted to acetic acid by oxygen
by the following reaction: C2H5OH + O2 CH3COOH + H2O. A 100 L
bottle of wine, labeled as 8.5% (by volume) ethanol, is found to have a
defective seal. Analysis of 1.00 mL showed that there were 0.0274 grams
of
acetic acid in that 1.00 mL. The density of ethanol is 0.816 g/mL and the
density of water is 1.00 g/mL.
a. What mass of oxygen must have leaked into the bottle?
b. What is the percent yield for the conversion of ethanol to acetic acid if
O2 is in excess?​

Answer 1

Answer:

Souring of wine occurs when ethanol is converted to acetic acid by oxygen by the following reaction: C2H5OH + O2 ---> CH3COOH + H2O. A 1.00 L bottle of wine, labeled as 8.5% (by volume) ethanol, is found to have a defective seal.

Explanation:

Answered: Souring of wine occurs when ethanol is…

QuestionAsked Feb 27, 20197783 views

Souring of wine occurs when ethanol is converted to acetic acid by oxygen by the following reaction:

C2H5OH + O2 --> CH3COOH + H2O

A 1.00 L bottle of wine, labeled as 8.5% (by volume) ethanol, is found to have a defective seal. Analysis of 1.00 mL showed that there were 0.0274 grams of acetic acid in that 1.00 mL. The density of ethanol is 0.816 g/mL and the density of water is 1.00 g/mL.

A. What mass of oxygen must have leaked into the bottle?

B. What is the percent yield for the conversion of ethanol to acetic acid if O2 is in excess?

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Expert Answer

Step 1

(a) The balanced equation of reaction between ethanol and oxygen is given as,

fullscreenStep 2

The mass of oxygen leaked into bottle is calculated as,

1ml of acetic acid = 0.0274g

1L of acetic acid = 27.4g

fullscreenStep 3

Thus, mass of oxygen leaked in...

Answer 2

Answer:

The mass of oxygen leaked into the bottle is 1460.16g and the percent yield is 30.3%.

Explanation:

Reaction of ethanol with oxygen gives acetic acid:

$C_2H_5OH + O_2\longrightarrow CH_3COOH + H_2O$

The mass of acetic acid in the 1.00 mL sample is 0.0274 g.

The mass of acetic acid in the 100 L bottle of wine is given

$\dfrac{0.0274 g}{1.00mL}\times 100L= \dfrac{0.0274 g}{1.00mL}\times 10^{5} mL=2740 g$

The molar mass of acetic acid is 60.052 g/mol.

The number of moles of acetic acid is

$\mbox{number of moles}=\dfrac{\mbox{mass}}{\mbox{molar mass}}= \dfrac{2740}{60.052}=45.63\mbox{mol}$

a. The number of moles of oxygen produced from 45.63 mol of acetic acid is

$\dfrac{1\mbox{mol} ~O_2}{1\mbox{mol} ~CH_3COOH}\times 45.63= 45.63\mbox{mol} ~O_2$

The molar mass of oxygen is 32.0 g/mol.

Thus, the mass of oxygen required to produce 2740 g of acetic acid is

$\mbox{mass}=\mbox{number of moles}\times {\mbox{molar mass}$

        $= 45.63\mbox{mol} ~O_2\times32.0 g/mol=1460.16g$

The mass of oxygen leaked into the bottle is 1460.16g.

b. Given the percentage of ethanol in the bottle is 8.5 %.

Volume of wine bottle is $100L=10^{5}mL$

Volume of ethanol in $10^{5}mL$ is

$=10^{5}\times \dfrac{8.5}{100} =8500mL$

The density of ethanol is 0.816 g/mL

Thus the mass of ethanol is

$density\times volume=0.816 \times8500=6936g$

The molar mass of ethanol is 46.07 g/mol.

The number of moles of ethanol is

$\mbox{number of moles}=\dfrac{\mbox{mass}}{\mbox{molar mass}}= \dfrac{6936}{46.07}=150.55\mbox{mol}$

The moles of acetic acid required to react with ethanol is

$\dfrac{1\mbox{mol} ~CH_3COOH}{1\mbox{mol} ~C_2H_5OH}\times 150.55= 150.55\mbox{mol}$

The mass of acetic acid is  

$\mbox{number of moles}\times {\mbox{molar mass}= 150.55\times60.052 =9040.83g$

Thus, the yield of the acetic acid in 100 L wine is 9040.83 g

The actual yield of acetic acid in 100 L wine is 2740 g

Therefore, the percent yield is

$\% yield=\dfrac{\mbox{Actual yield}}{\mbox{Theoretical yield}}\times 100$

           $= \dfrac{\mbox{2740}}{\mbox{9040.83}}\times 100=150.55\mbox{mol}=30.3\%$

Therefore, the percent yield for the conversion of ethanol to acetic acid with excess $O_2$ is 30.3%.

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