The alcoholic fermentation of350 g/l glucose syrup ceases when the alcoholic degree of the mixture attains 16º (16ml ethanol in 100ml of the solution)
Write the reaction of the fermentation reaction.
Consider the fermentation of 100L of the glucose syrup. Determine the volume and the mass of ethanol formed as fermentation ceases.
The fermentation of glucose can be represented by the following equation:
C6H12O6 → 2 C2H5OH + 2 CO2
This equation shows that one molecule of glucose (C6H12O6) is converted into two molecules of ethanol (C2H5OH) and two molecules of carbon dioxide (CO2).
To determine the volume and mass of ethanol formed, we first need to calculate the moles of glucose in 100 L of the glucose syrup. The molar mass of glucose (C6H12O6) is:
6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol
The number of moles of glucose can be calculated using the molar mass and the given mass of glucose syrup:
moles of glucose = mass of glucose syrup / molar mass of glucose
= (350 g/L) / (180.18 g/mol)
= 1.942 mol/L
Since we have 100 L of the glucose syrup, the total moles of glucose are:
total moles of glucose = moles of glucose × volume
= 1.942 mol/L × 100 L
= 194.2 mol
According to the balanced equation, each mole of glucose yields 2 moles of ethanol. Therefore, the total moles of ethanol produced are:
moles of ethanol = 2 × moles of glucose
= 2 × 194.2 mol
= 388.4 mol
To determine the volume of ethanol formed, we need to convert the moles of ethanol to volume using the given alcoholic degree of the mixture. We know that 16 mL of ethanol is present in 100 mL of the solution.
Therefore, the volume of ethanol formed is:
volume of ethanol = (moles of ethanol / total moles) × volume of solution
= (388.4 mol / 194.2 mol) × 100 L
= 200 L
Finally, to determine the mass of ethanol formed, we can use the molar mass of ethanol (C2H5OH), which is:
2(12.01 g/mol) + 6(1.01 g/mol) + 16.00 g/mol = 46.07 g/mol
The mass of ethanol formed is:
mass of ethanol = moles of ethanol × molar mass of ethanol
= 388.4 mol × 46.07 g/mol
= 17,911.65 g
≈ 17.91 kg
Therefore, the volume of ethanol formed is 200 L, and the mass of ethanol formed is approximately 17.91 kg.
The reaction for alcoholic fermentation can be represented by the following equation:
C6H12O6 (glucose) → 2 C2H5OH (ethanol) + 2 CO2 (carbon dioxide)
Now, let's calculate the volume and mass of ethanol formed during the fermentation process.
Given:
- Concentration of glucose syrup: 350 g/L
- Alcoholic degree (volume of ethanol in 100 ml of solution): 16º
Step 1: Determine the number of moles of glucose in 100 L of glucose syrup.
To find the number of moles, we divide the mass of glucose by its molar mass:
Molar mass of glucose (C6H12O6) = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol
Number of moles of glucose = (350 g/L) / (180.18 g/mol) = 1.9429 mol/L
Number of moles of glucose in 100 L = (1.9429 mol/L) * 100 L = 194.29 mol
Step 2: Determine the number of moles of ethanol formed.
From the balanced equation, we know that 1 mole of glucose yields 2 moles of ethanol.
So, the number of moles of ethanol formed = 2 * (number of moles of glucose) = 2 * 194.29 mol = 388.58 mol
Step 3: Calculate the volume of ethanol formed.
Using the given alcoholic degree of 16º, we can calculate the volume of ethanol formed as follows:
Volume of ethanol = (16 ml ethanol / 100 ml of solution) * 100 L = 16 L
Step 4: Calculate the mass of ethanol formed.
To determine the mass of ethanol, we need to multiply the number of moles of ethanol by its molar mass:
Molar mass of ethanol (C2H5OH) = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.08 g/mol
Mass of ethanol formed = (number of moles of ethanol) * (molar mass of ethanol)
= 388.58 mol * 46.08 g/mol = 17,906.35 g
Therefore, as fermentation ceases, the volume of ethanol formed is 16 L, and the mass of ethanol formed is approximately 17,906.35 g.