C6H12O6(aq)->2 C2H5OH(aq) + 2CO2(g)
Fermentation of 744mL grape juice(density=1.0 g/cm^3) is allowed to take place in a bottle witha V of 825mL until 12% by volume is ethanol (C2H5OH). Assuming CO2 is insoluble in H2O what would the pressure be of the CO2 inside the wine bottle at 34C (density of ethanol is .79g/cm^3
Write the balanced equation.
Convert mL grape juice to grams using density given and convert that to mols.
Convert 12% by volume of ethanol to grams, then to mols,
Convert mols ethanol produced to mols CO2 using the coefficients in the equation.
Use PV = nRT to calculate the pressure.
n is the # mols CO2 calculated above. Don't forget to change T to Kelvin. Note: You may need to correct for the volume. The empty space above the level of the grape juice starts out at 825 mL - 744 mL but it will be different at the end since ethanol has a different density than grape juice.
Post your work if you need additional assistance.
C6H12O6(aq)--2C2H5Oh(aq)+2CO2(g)
To find the pressure of CO2 inside the wine bottle, we can use the Ideal Gas Law equation, which states:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
First, we need to determine the number of moles of CO2 produced during fermentation. According to the balanced equation, for every 1 mole of C6H12O6 (glucose), 2 moles of C2H5OH (ethanol) and 2 moles of CO2 are produced.
To calculate the number of moles of C2H5OH produced, we need to know the concentration of ethanol in the grape juice. The given information states that the fermentation is allowed to take place until 12% by volume is ethanol. Since the density of grape juice is 1.0 g/cm^3, we can assume that 1 mL is approximately equal to 1 gram.
So, if we have 744 mL of grape juice, the mass of ethanol produced is 0.12 * 744 g = 89.28 g.
Now, let's find the number of moles of ethanol:
n(C2H5OH) = mass / molar mass
where the molar mass of C2H5OH is 46.07 g/mol.
n(C2H5OH) = 89.28 g / 46.07 g/mol ≈ 1.94 mol
Since we have 825 mL of total volume in the bottle, and we know that 12% by volume is ethanol, we can calculate the volume of ethanol in milliliters:
V(C2H5OH) = 0.12 * 825 mL = 99 mL
Since the density of ethanol is given as 0.79 g/cm^3, we can convert this volume of ethanol to grams:
mass(C2H5OH) = volume * density = 99 mL * 0.79 g/cm^3 ≈ 78.21 g
Now, let's find the number of moles of CO2 produced using the stoichiometry of the reaction:
Since 2 moles of ethanol are produced, 2 moles of CO2 will also be produced.
n(CO2) = 2 * n(C2H5OH) = 2 * 1.94 mol = 3.88 mol
Now that we have the number of moles of CO2, we can use the Ideal Gas Law equation. But before we do that, we need to convert the temperature from degrees Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 34°C + 273.15 = 307.15 K
Now we can plug the values into the equation:
PV = nRT
Rearranging the equation to solve for P:
P = nRT / V
P = (3.88 mol) * (0.0821 L·atm/(mol·K)) * (307.15 K) / 825 mL
Note that the volume must be converted from milliliters (mL) to liters (L) since the R constant is in terms of liters.
P ≈ 0.468 atm
Therefore, the pressure of CO2 inside the wine bottle would be approximately 0.468 atm at 34°C.