100 grams of glucose is mixed with 500 grams of water. 10 grams of yeast is then added to the mixture. If the reaction goes to completion, what is the mass of ethanol in the final mixture?
Here is a solved example of a stoichiometry problem. Follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the mass of ethanol produced in the final mixture, we need to calculate the amount of glucose that is consumed by yeast during fermentation. The equation for the fermentation of glucose by yeast is:
C6H12O6 (glucose) → 2C2H5OH (ethanol) + 2CO2 (carbon dioxide)
From the balanced equation, we can see that 1 mole of glucose reacts to produce 2 moles of ethanol. To calculate the moles of glucose present in the mixture, we need to convert its mass to moles using its molar mass.
The molar mass of glucose (C6H12O6) is:
(6 x atomic mass of carbon) + (12 x atomic mass of hydrogen) + (6 x atomic mass of oxygen)
= (6 x 12.01 g/mol) + (12 x 1.01 g/mol) + (6 x 16.00 g/mol)
= 180.18 g/mol
We can now calculate the number of moles of glucose by dividing its mass by its molar mass:
moles of glucose = mass of glucose / molar mass of glucose
= 100 g / 180.18 g/mol
≈ 0.554 mol
According to the balanced equation, 1 mole of glucose produces 2 moles of ethanol. Therefore, the moles of ethanol produced will be twice the number of moles of glucose:
moles of ethanol = 2 x moles of glucose
= 2 x 0.554 mol
≈ 1.108 mol
Now, to find the mass of ethanol produced, we multiply the moles of ethanol by its molar mass:
mass of ethanol = moles of ethanol x molar mass of ethanol
= 1.108 mol x (2 x atomic mass of carbon) + (6 x atomic mass of hydrogen) + (1 x atomic mass of oxygen)
= 1.108 mol x (2 x 12.01 g/mol + 6 x 1.01 g/mol + 1 x 16.00 g/mol)
≈ 105.50 g
Therefore, the mass of ethanol in the final mixture is approximately 105.50 grams.