During the fermentation of glucose, 118 liters of gas were released. How much glucose has decomposed?

1. Let’s write down the formula of glucose fermentation:

C6H12O6 = 2CO2 + 2C2H5OH (alcohol and carbon dioxide are formed).

2. From Avogadro’s law it follows that 1 mole of any gas under normal conditions takes a volume equal to 22, 4 liters.

Let’s find the amount of released CO2: 118 l / 22.4 l / mol = 5.3 mol.

3. Let’s make the proportion:

1 mol C6H12O6 = 2 mol CO2

x mol = 5.3 mol CO2

x = 2.65 mol

Mass of 1 mol of glucose = 180 g.

Then the mass of consumed glucose = 180 g / mol * 2.65 mol = 477 g.

Answer: 477