Question - chem.

13 - In a fermentation reaction glucose is converted to alcohol and carbon dioxide according to the following equation

C6H12O6 → 2C2H5OH + 2CO2

What mass of alcohol and what volume of carbon dioxide would be produced from 109 grams of glucose?

Number of moles of glucose = mass/RMM = 109 / ((6x12)+(12x1)+(6x16)) = 0.60555moles

Moles of alcohol = moles of carbon dioxide = 0.60555555 x 2 = 1.21111 moles

Mass of alcohol = 1.21111 x ((12x2)+(1x5)+16+1) = 55.7g

Mass of carbon dioxide = 1.21111 x (12+(2x16)) = 53.3g

Assuming this is at room temperature and pressure pV = nRT, so V = nRT/P --> V = (1.21111 x 8.31 x 298) / (1.01 x 10^5)

So the volume of carbon dioxide produced is 0.0297 m^3

Check by conserving mass:
Mass on the left hand side = 109g
Mass on the right hand side = 55.7+53.3 = 109g

Hence the solution is plausible.

PhD.