Question

In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is the following. C6H12O6(aq) → 2 C2H5OH(aq) + 2 CO2(g) Fermentation of 754 mL of grape juice (density = 1.0 g/cm3) is allowed to take place in a bottle with a total volume of 840. mL until 13% by volume is ethanol (C2H5OH). Assuming that the CO2 is insoluble in H2O (actually a wrong assumption), what would be the pressure of CO2 inside the wine bottle at 21°C? (The density of ethanol is 0.79 g/cm3.) WebAssign will check your answer for the correct number of significant figures. atm

Answer 1

Answer:

The pressure inside the wine bottle at 21 °C is 4.8 · 10² atm

Explanation:

Hi there!

We know that 1 mol of CO₂ is produced per mol of produced ethanol.

If the final concentration of ethanol is 13%, let´s calculate how many moles of ethanol are present at that concentration.

A concentration of 13% means that in 100 ml of solution, 13 ml is dissolved ethanol. We have 754 ml of solution, then, the volume of ethanol will be:

754 ml solution · (13 ml ethanol/100 ml solution) = 98 ml ethanol

With the density, we can calculate the mass of ethanol present:

density = mass/ volume

0.79 g/ml = mass / 98 ml

mass = 0.79 g/ml · 98 ml

mass = 77 g

The molar mass of ethanol is 46.07 g/mol, then 77 g of ethanol is equal to:

77 g · (1 mol/46.07 g) = 1.7 mol

Then, the number of moles of CO₂ produced will be 1.7 mol.

Using the equation of the ideal gas law, we can calculate the pressure of CO₂:

P = nRT/V

Where:

P = pressure

n = number of moles

R = ideal gas constant

T = temperature

V = volume

The volume will be the headspace of the bottle (840 ml - 754 ml) 86 ml = 0.086 l.

The temperature in kelvin will be: 21 + 273 = 294 K

The gas constant is 0.082 l atm / K mol

Then:

P = (1.7 mol · 0.082 l atm/K mol · 294 K)/ 0.086 l

P = 4.8 · 10² atm

The pressure inside the wine bottle at 21 °C is 4.8 · 10² atm