Question

Consider the fermentation reaction of glucose: C6H12O6 ----) 2C2H5OH+2CO2

A 1.00-mol sample of C6H12O6 was placed in a vat with excess yeast. If 35.1 g of C2H5OH was obtained, what was the percent yield of C2H5OH?

Answer 1

1 mole of C6H12O6=2 moles of C2H5OH


moles of C2H5OH*(46.06867g of C2H5OH/mole)=g of C2H5OH

(g of C2H5OH/35.1 g of C2H5OH)*100= percent yield of C2H5OH

Answer should have three significant figures.

Answer 2

Opps, typo.


Last part should say (35.1 g of C2H5OH/g of C2H5OH)*100= percent yield of C2H5OH

Answer should have three significant figures


***** Note: g of C2H5OH=46.1g

Answer 3

COULD YOU PLS EXPLAIN IT TO ME.

Answer 4

Okay, looking at the reaction that you provided,

1 mole of C6H12O6=2 moles of C2H5OH

You placed one mole of C6H12O6 in the vat. Your reaction tells you that you will get 2 moles of C2H5OH for every 1 mole of C6H12O6.

C6H12O6 ----> 2C2H5OH+2CO2

We know that 1 mole of a substance=molecular mass, so two moles of C2H5OH= 92.2g of C2H5OH produced in theory.

The precent yield=(actual yield/theoretical yield)*100

The precent yield=(35.1g/92.2g)*100=38.1%


****Sorry about the note in my last post; Forgot that it was a 1 to 2 mole ratio.

Answer 5

fermentation is chemical decomposition in which glucose is converted into ethyl alcohol and carbon dioxide then what will be the amount of ethyl alcohol in grams and moles, which can be obtained by fermentation of 5000 grams of glucose.