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

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Description

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?

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Explanation & Answer

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

To answer these questions you need to use stoichiometry of the balanced equation given.

a) .7 mols of glucose * (2mols CO2/1 mol glucose) = 1.4 mols of CO2

b) for this question we will need the MM of glucose and alcohol.  46.07 g/mol for ethyl al, and 180.16 g/mol for glucose. 

7.5 g of ethyl al. * (1 mol ethyl alcohol/46.07) * (1 mol glucose/2 mols ethyl al) * (180.16 g glucose/mol) =  14.66 g of glucose

c) is similar to b, but wants the weight of CO2 instead of glucose.

7.5 g of ethyl al. * (1 mol ethyl alcohol/46.07) * (2 mol CO2/2 mols ethyl al) * (44.01g CO2/mol) =  7.16 g of CO2

Please let me know if you didn't understand something that I did or need any clarification. I'm always happy to answer your questions!


Anonymous
Excellent resource! Really helped me get the gist of things.

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