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

Chemistry
Tutor: None Selected Time limit: 1 Day

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

C6H12O6(aq) → 2 C2H5OH(aq) + 2 CO2(g)

a) How many moles of CO2 are produced when 0.700 mol of C6H12O6 reacts in this fashion?

b) How many grams of C6H12O6 are needed to form 7.50 g of C2H5OH? 

c)  How many grams of CO2 form when 7.50 g of C2H5OH are produced?

Jul 24th, 2015

Thank you for the opportunity to help you with your question!

a)DATA:

Moles of glucose= 0.700mol

Moles of CO2 produced=?

SOLUTION:

1C6H12O6 = 2CO2

0.700mol C6H12O6 = (2*0.700)/1

  = 1.4mol CO2

b) DATA:

mass of ethyl alcohol = 7.50g

mass of glucose =?

SOLUTION:

Moles of ethyl alcohol= mass of ethyl alcohol/molecular mass of ethyl alcohol

  = 7.50/46

Moles of ethyl alcohol = 0.1630mol of ethyl alcohol

2 C2H5OH =1C6H12O6

0.1630 mol of C2H5OH = (1*0.1630)/2

  = 0.0815 mol of C6H12O6 

Mass of glucose = moles of glucose * molecular mass of glucose

  = 0.0815*180

  = 14.67g 

c) DATA:

mass of ethyl alcohol = 7.50g

mass of CO2 produced=?

SOLUTION:
Moles of ethyl alcohol= mass of ethyl alcohol/molecular mass of ethyl alcohol

  = 7.50/46

Moles of ethyl alcohol = 0.1630mol of ethyl alcohol

2 C2H5OH =1C6H12O6

0.1630 mol of C2H5OH = (1*0.1630)/2

  = 0.0815 mol of C6H12O6 

1C6H12O6 = 2 CO2

0.0815 mol C6H12O6  = (2*0.0815)/1

  = 0.163mol CO2

Mass of CO2 = moles of CO2*molecular mass of carbondioxide

  = 0.163 *44

  = 7.172g


Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 24th, 2015

Thank you so much, but one last thing.. I would like the answer to a in 3 sig figs so is there a more precise version or shall I use 1.40 moles?

Jul 24th, 2015

you can write 1.40mole

Jul 24th, 2015

Your too good :)

Jul 24th, 2015

thank you... :)

Jul 24th, 2015

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