Chem yield help
C6H12O6 → 2C2H5OH + 2CO2
in an experiment, 268 g of ethanol (Mr = 46.0) were made from 1.44 kg of glucose (Mr = 180.0).
What is the percentage yield?
Answer ; 36.4
How? Can someone explain please?
in an experiment, 268 g of ethanol (Mr = 46.0) were made from 1.44 kg of glucose (Mr = 180.0).
What is the percentage yield?
Answer ; 36.4
How? Can someone explain please?
Original post by studybloomer
C6H12O6 → 2C2H5OH + 2CO2
in an experiment, 268 g of ethanol (Mr = 46.0) were made from 1.44 kg of glucose (Mr = 180.0).
What is the percentage yield?
Answer ; 36.4
How? Can someone explain please?
in an experiment, 268 g of ethanol (Mr = 46.0) were made from 1.44 kg of glucose (Mr = 180.0).
What is the percentage yield?
Answer ; 36.4
How? Can someone explain please?
1440g/180=8moles, 1/2 ratio so 8x2=16, mass=m x mr so 16x46 = 736, 268/736 x100= 36.4%
Original post by studybloomer
C6H12O6 → 2C2H5OH + 2CO2
in an experiment, 268 g of ethanol (Mr = 46.0) were made from 1.44 kg of glucose (Mr = 180.0).
What is the percentage yield?
Answer ; 36.4
How? Can someone explain please?
in an experiment, 268 g of ethanol (Mr = 46.0) were made from 1.44 kg of glucose (Mr = 180.0).
What is the percentage yield?
Answer ; 36.4
How? Can someone explain please?
Original post by Bwoah
1440g/180=8moles, 1/2 ratio so 8x2=16, mass=m x mr so 16x46 = 736, 268/736 x100= 36.4%
how
Reply 4
Original post by studybloomer
C6H12O6 → 2C2H5OH + 2CO2
in an experiment, 268 g of ethanol (Mr = 46.0) were made from 1.44 kg of glucose (Mr = 180.0).
What is the percentage yield?
Answer ; 36.4
How? Can someone explain please?
in an experiment, 268 g of ethanol (Mr = 46.0) were made from 1.44 kg of glucose (Mr = 180.0).
What is the percentage yield?
Answer ; 36.4
How? Can someone explain please?
There are four values you need to consider:
•
The mass of glucose used in the experiment is 1.44 kg = 1440 g.
•
The mass of ethanol produced is given as 268 g.
•
Ethanol (Mr = 46.0 g/mol)
•
Glucose (Mr = 180.0 g/mol)
moles of glucose = mass of glucose / molecular weight of glucose = 1440 g / 180.0 g/mol = 8.0 mol
Now, you have to find the moles of ethanol:
moles of ethanol = 2 × moles of glucose = 2 × 8.0 mol = 16.0 mol; because molar ratio is 1:2 as per
C6H12O6 → 2C2H5OH
Finally, you can calculate the theoretical yield of ethanol:
theoretical yield of ethanol = 16.0 mol × 46.0 g/mol = 736.0 g
The actual yield of ethanol is 268 g. To find the percentage yield, you can use the formula:
% yield = (actual yield / theoretical yield) × 100 = (268 g / 736.0 g) × 100 = 36.4%
Kind regards from Italy!
Sandro
(edited 7 months ago)
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