Academic Subjects
➤
Chemistry
➤
General Chemistry
➤
Chemical Reactions
➤
Theoretical Yield Of Ethanol
Determine the theoretical yield in grams of ethanol when 833 g of glucose is used according to the following fermentation reaction
C
6
H
12
O
6(aq)
⇒
2 C
2
H
5
OH
(l)
+ 2 CO
2(g)
We already have our balanced equation, so let's find the molecular weight of glucose and ethanol
Glucose C
6
H
12
O
6
C = 12.01 g/mol (6) = 72.06 g/mol
H = 1.01 g/mol (12) = 12.12 g/mol
O = 16.00 g/mol (6) = 96.00 g/mol
180.18 g/mol
EtOH C
2
H
5
OH
C = 12.01 g/mol (2) = 24.02 g/mol
H = 1.01 g/mol (6) = 6.06 g/mol
O = 16.00 g/mol (1) = 16.00 g/mol
46.08 g/mol
And now we convert g C
6
H
12
O
6
into g C
2
H
5
OH
833
g C
6
H
12
O
6
(1
mole C
6
H
12
O
6
)
/
(180.18
g C
6
H
12
O
6
)
(2
moles C
2
H
5
OH
)
/
(1
mole C
6
H
12
O
6
)
(46.08 g C
2
H
5
OH)
/
(1
mole C
2
H
5
OH
)
= 426.0699 g C
2
H
5
OH
The theoretical yield of EtOH is 426 g