If 675.0 g of glucose yields 106.0 mL of ethanol, what is the percent yield for the reaction?

Answer 1

To do this problem you need two pieces of information: the balanced equation and the density of ethanol (which I looked up: it is 0.7892 g/mL @ 20^oC, and I will use this value.)

C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂ The molar masses of glucose & ethanol are, respectively, ~180.0 g/mol & 46.00 g/mol) For simplicity let Et = ethanol & Glu = glucose

(1 ml /0.7892 g Et) (46.00 g Et) (46.00 g Et/1 mol Et) (2 mol Et/1 mol Glu) (1 mol Glu/180.0 g Glu) 675.0 g Glu = 437.2 mL of Et

This is the theoretical yield. The actual yield is given as 106.0 mL. Therefore, (AY/TY) x 100 = 24.25%