Answer:
9.221 g
Step-by-step explanation:
To calculate the mass of ethanol (C2H5OH) obtained from 36g of glucose (C6H12O6), we first need to understand the chemical reaction that converts glucose into ethanol. The reaction is called fermentation, and it can be represented as follows:
C6H12O6 (glucose) → 2 C2H5OH (ethanol) + 2 CO2 (carbon dioxide)
In this reaction, one molecule of glucose is converted into two molecules of ethanol and two molecules of carbon dioxide.
Now, let's calculate the molar masses:
Molar mass of glucose (C6H12O6):
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of glucose (C6H12O6) = (6 * C) + (12 * H) + (6 * O) = 6 * 12.01 + 12 * 1.01 + 6 * 16.00 = 72.06 + 12.12 + 96.00 = 180.18 g/mol
Molar mass of ethanol (C2H5OH):
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of ethanol (C2H5OH) = (2 * C) + (6 * H) + (1 * O) = 2 * 12.01 + 6 * 1.01 + 1 * 16.00 = 24.02 + 6.06 + 16.00 = 46.08 g/mol
Now, we can set up a proportion to find the mass of ethanol produced from 36g of glucose:
(36 g of glucose) / (180.18 g/mol) = (x g of ethanol) / (46.08 g/mol)
Solve for "x" (mass of ethanol):
x = (36 g of glucose) * (46.08 g/mol) / (180.18 g/mol)
x ≈ 9.221 g
So, approximately 9.221 grams of ethanol will be obtained from 36 grams of glucose during the fermentation process.