manzk1k
Answered

The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2:


C6H12O6(aq) → 2 C2H5OH(aq) + 2 CO2(g)

AHow many moles of CO2 are produced when 0.300 mol of C6H12O6 reacts in this fashion?

BHow many grams of C6H12O6 are needed to form 2.00 g of C2H5OH?

CHow many grams of CO2 form when 2.00 g of C2H5OH are produced?

Verified Answer :

alexdagrape

To answer your question I will use dimensional analysis, which is used by cancelling out the units. I will also use the balanced equation provided as a conversion factor.

A) First start out with the 0.300 mol of C6H12O6...
0.300 mol C6H12O6 (2 mol CO2 / 1 mol C6H12O6) = 0.600 mol CO2

The significant figures (sig figs) at still three, the 2 is a conversion counting number and does not count

B) First change 2.00 g of C2H5OH to moles of C2H5OH...
The molecular mass of C2H5OH is...
2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol
This can be used as a conversion factor to change grams to moles. 
2.00 g C2H5OH (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH

Second, you can change the moles of C2H5OH to moles of C6H12O6..
0.0434 mol C2H5OH (1 mol C6H12O6 / 2 mol C6H12O6) = 0.0217 mol C6H12O6

Third, change moles of C6H12O6 to grams...
MM = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = 180.16 g/mol
0.0217 mol C6H12O6 (180.16 g C6H12O6 / 1 mol C6H12O6) = 3.91 g C6H12O6

C) Now I am going to put it all into one long dimensional analysis problem.
MM of CO2 = 44.01 g/mol
MM of C2H5OH = 46.07 g/mol

2.00 g C2H5OH (1 mol C2H5OH / 46.07 g C2H5OH) (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2

I hope this helped and I am sorry that I talked to much, I just didn't want to miss anything!

khuranashilpa76

A)  0.300 mol of C6H12O6...

0.300 mol C6H12O6 (2 mol CO2 / 1 mol C6H12O6) = 0.600 mol CO2

B) 2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol

2.00 g C2H5OH (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH

C) MM of CO2 = 44.01 g/mol

MM of C2H5OH = 46.07 g/mol

2.00 g C2H5OH (1 mol C2H5OH / 46.07 g C2H5OH) (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2

What are the products of the fermentation of glucose?

The fermentation products from glucose were butyrate, acetate, and formate; those from lactate were 1-butanol, ethanol, butyrate, and formate.

What causes glucose to ferment?

When yeast is added it feeds on the sugar in the absence of oxygen to form wine (a solution of ethanol) and carbon dioxide. A chemical reaction called fermentation takes place in which the glucose is broken down to ethanol by the action of enzymes in the yeast.

Learn more about glucose here: https://brainly.com/question/461228

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