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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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
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