The following equation shows the conversion of ethanol c2h5oh in a wine sample into acetic acid c2h4o2 through the action of oxygen from air c2h5oh+o2-> c2h4o2+h2o this reaction drastically changes the taste of the wine. if 100.0ml of wine initially contained 12.0g of ethanol, and after a period of time 4.00g of acetic acid were detected, what percent of the ethanol had been converted?

Given:

Volume of wine = 100 ml

Initial Mass of ethanol = 12.0 g

Mass of acetic acid formed = 4.00 g

To determine:

% ethanol converted to acetic acid

Explanation:

Molar mass of C2H5OH (ethanol) = 46 g/mol

Molar mass of C2H4O2 (acetic acid) = 60 g/mol

The reaction of ethanol in air is:

C2H5OH + O2 → C2H4O2 + H2O

Based on the stoichiometry: 1 mole of ethanol produces 1 mole of acetic acid

Moles of ethanol initially present = mass of ethanol/molar mass

= 12/46 = 0.2609 moles

Moles of acetic acid produced = moles of ethanol converted

= 4/60 = 0.0667 moles

Mass of ethanol converted = Moles of ethanol * molar mass

= 0.0667 * 46 = 3.07 g

% ethanol converted = (mass of ethanol converted/initial mass )*100

% converted = (3.07/12.0) *100 = 25.6%

Ans: 25.6% of ethanol gets converted to acetic acid

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm