The formation of ethyl alcohol (C2H5OH) by the fermentation of glucose (C6H12O6) may be represented by:
C6H12O6 2C2H5OH + 2CO2
If a particular glucose fermentation process is 87.0% efficient, how many grams of glucose would be required for the production of 51.0 g of ethyl alcohol (C2H5OH)?
(a) 68.3 g
(b) 75.1 g
(c) 115 g
(d) 229 g
(e) 167 g
The reaction coefficients can be related by the unit of mols, which you get from the molar mass.
51g C6H12O6×1 mol glucose
180.156g glucose = 0.1388 mols glucose
If you had 1 mol of glucose, you would get 2 mols ethanol
. So since you have 0.1388 mols
of glucose, you get 0.2775 mols ethanol
.
0.1388 mols glucose × 2 mols ethanol
1 mol glucose =
0.2775 mols ethanol
And this many mols has a mass of...
0.2775 mols CH3CH2OH × 51 g
1 mol EtOH =
115g ethanol
Comments
How did you get the 0.1388?
how did you get 0.1388?
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