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?
Thank you for the opportunity to help you with your question!
Moles of glucose= 0.700mol
Moles of CO2 produced=?
1C6H12O6 = 2CO2
0.700mol C6H12O6 =
= 1.4mol CO2
mass of ethyl alcohol = 7.50g
mass of glucose =?
Moles of ethyl alcohol= mass of ethyl alcohol/molecular mass
of ethyl alcohol
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
mass of ethyl alcohol
mass of CO2 produced=?
Moles of ethyl alcohol= mass of ethyl alcohol/molecular mass of ethyl alcohol
1C6H12O6 = 2 CO2
0.0815 mol C6H12O6 = (2*0.0815)/1
Mass of CO2 = moles of CO2*molecular
mass of carbondioxide
= 0.163 *44
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?
you can write 1.40mole
Your too good :)
thank you... :)
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