I need help on this stoichiometry problem. I don't know whay my answer is incorrect.
The fermentation of sugar to produce ethyl alcohol occurs by the following reaction:
C6H12O6 -- 2C2H5OH + 2CO2
What mass of ethyl alcohol can be made from 1.00 kg of sugar?
I keep getting 1012.32 g, but the solution says 511g. please help!
And 511 it is.
moles sugar = 1000g/molar mass sugar = ??
moles ethyl alcohol produced = 2 x moles sugar.
mass ethyl alcohol = moles x molar mass.
To solve this stoichiometry problem, you will need to follow a few steps:
Step 1: Write down the balanced equation.
The given balanced equation is: C6H12O6 ⟶ 2C2H5OH + 2CO2
Step 2: Determine the molar mass of the substance you are given.
The molar mass of C6H12O6 (sugar) is calculated as follows:
(6 x 12.01 g/mol) + (12 x 1.01 g/mol) + (6 x 16.00 g/mol) = 180.18 g/mol
Step 3: Use the molar mass to convert the given mass of sugar into moles.
Given: 1.00 kg of sugar = 1000 g
Moles of sugar = 1000 g / 180.18 g/mol = 5.550 mol
Step 4: Apply the stoichiometry relationship to determine the moles of ethyl alcohol produced.
According to the equation, 1 mole of sugar produces 2 moles of C2H5OH.
Moles of C2H5OH = 5.550 mol x (2 mol C2H5OH / 1 mol of sugar) = 11.100 mol of C2H5OH
Step 5: Convert the moles of ethyl alcohol into grams.
To calculate the mass of the ethyl alcohol, we need to multiply the moles of C2H5OH by its molar mass.
The molar mass of C2H5OH (ethyl alcohol) is calculated as follows:
(2 x 12.01 g/mol) + (5 x 1.01 g/mol) + (1 x 16.00 g/mol) = 46.07 g/mol
Mass of C2H5OH = 11.100 mol x 46.07 g/mol = 511.77 g
Therefore, the correct mass of ethyl alcohol that can be made from 1.00 kg of sugar is 511.77 g, which can be rounded to 511 g.