How many grams of glucose have undergone fermentation if 450 dm3 of CO2 has been formed.

Let’s write down the reaction equations:
C6H12O6 = 2C2H5OH + 2CO2 ↑.
It can be seen from the reaction equation that:
ν (CO2) / 2 = ν (C6H12O6).
m (C6H12O6) / M (C6H12O6) = V (CO2) / Vm (CO2).
Vm (CO2) = 22.4 L / mol.
Determine the molar mass of glucose:
M (C6H12O6) = 12 * 6 + 1 * 12 + 16 + 6 = 180 g / mol
Determine the mass of glucose:
m (C6H12O6) = V (CO2) * M (C6H12O6) / (2 * Vm (CO2)).
Substituting the numerical values, we find the mass of glucose:
m (C6H12O6) = V (CO2) * M (C6H12O6) / (2 * Vm (CO2)) = 450 * 180 / (2 * 22.4) = 1808 g.
Answer: the mass of glucose that entered into fermentation in 1808.