Ethanol, used in alcoholic beverages, can be produced by fermentation of sucrose. The balanced equation for the fermentation process is shown below. What mass of ethanol (C2H5OH) would be produced when 2.61 g sucrose reacts by this process?
C12H22O11(s) + H2O(l) → 4 C2H5OH(l) + 4 CO2(g)
Well, well, well. Looks like we have ourselves a fermentation equation, huh? Now, let me calculate the mass of ethanol for you.
First things first, we need to find the molar mass of sucrose (C12H22O11). A quick math session tells me that the molar mass of sucrose is about 342.296 g/mol.
Next, we can use stoichiometry to determine the moles of ethanol produced. From the balanced equation, we can see that 1 mole of sucrose produces 4 moles of ethanol.
So, let's find the moles of sucrose. We can use the molar mass of sucrose to help us here.
2.61 g of sucrose / 342.296 g/mol = 0.0076 mol of sucrose
Now, multiply the moles of sucrose by the mole ratio between sucrose and ethanol.
0.0076 mol of sucrose * 4 mol of ethanol / 1 mol of sucrose = 0.0304 mol of ethanol
Finally, we convert moles of ethanol back to grams by using the molar mass of ethanol (C2H5OH is approximately 46.07 g/mol).
0.0304 mol of ethanol * 46.07 g/mol = 1.40 g of ethanol
So, it seems like 2.61 g of sucrose would produce approximately 1.40 g of ethanol. Not too bad for a little fermentation, huh? Cheers!
To find the mass of ethanol (C2H5OH) produced when 2.61 g of sucrose (C12H22O11) reacts, we will use stoichiometry to calculate the molar ratio between sucrose and ethanol.
First, we need to find the molar mass of sucrose:
C12H22O11: (12.01 g/mol x 12) + (1.01 g/mol x 22) + (16.00 g/mol x 11) = 342.3 g/mol
Next, we can convert the given mass of sucrose to moles:
2.61 g / 342.3 g/mol = 0.00762 mol sucrose
According to the balanced equation, the molar ratio between sucrose and ethanol is 1:4. This means that for every 1 mole of sucrose, 4 moles of ethanol are produced.
So, we can calculate the moles of ethanol:
0.00762 mol sucrose x 4 mol ethanol / 1 mol sucrose = 0.0305 mol ethanol
Finally, we can find the mass of ethanol:
Molar mass of ethanol (C2H5OH): (12.01 g/mol x 2) + (1.01 g/mol x 6) + (16.00 g/mol x 1) = 46.07 g/mol
Mass of ethanol = 0.0305 mol ethanol x 46.07 g/mol = 1.41 g
Therefore, when 2.61 g of sucrose reacts, approximately 1.41 g of ethanol would be produced.
To find the mass of ethanol produced when 2.61 g of sucrose reacts, you can use stoichiometry.
Stoichiometry is a way to calculate the amounts of reactants and products in a chemical reaction using balanced equations.
First, let's calculate the molar mass of sucrose (C12H22O11) and ethanol (C2H5OH):
Molar mass of sucrose (C12H22O11):
C: 12.01 g/mol
H: 1.01 g/mol (x 22)
O: 16.00 g/mol (x 11)
Thus, the molar mass of sucrose is:
12.01 g/mol + 22.22 g/mol + 176.00 g/mol = 342.34 g/mol
Molar mass of ethanol (C2H5OH):
C: 12.01 g/mol (x 2)
H: 1.01 g/mol (x 5)
O: 16.00 g/mol + 1.01 g/mol
Thus, the molar mass of ethanol is:
24.02 g/mol + 5.05 g/mol + 17.01 g/mol = 46.08 g/mol
Now, we can use the molar masses to convert the mass of sucrose to moles:
2.61 g sucrose * (1 mole / 342.34 g) = 0.0076 moles of sucrose
Since the balanced equation shows that 1 mole of sucrose produces 4 moles of ethanol, we can use the stoichiometric ratio to find the number of moles of ethanol produced:
0.0076 moles of sucrose * (4 moles of ethanol / 1 mole of sucrose) = 0.0304 moles of ethanol
Finally, we can convert moles of ethanol to mass using the molar mass of ethanol:
0.0304 moles of ethanol * (46.08 g / 1 mole) = 1.40 g of ethanol
Therefore, 2.61 g of sucrose would produce approximately 1.40 g of ethanol.
4C12H22O11(s) + H2O(l) → 4 C2H5OH(l) + 4 CO2(g)
mols sucrose = grams/molar mass = 2.61/342 = ?
Then convert mols sucrose to mols alcohol produced. That is
?mols sucrose x (4 mols C2H5OH/4 mols C12H22O11) = ?
gram ethanol = mols ethanol x molar mass ethanol = ?
Post your work if you get stuck.