Question

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

Answer 1

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