How many grams of glucose was fermented if 450 dm3 CO2 was formed?

Decision:
1. Let’s write down the reaction equation:
С6Н12О6 = 2С2Н5ОН + 2СО2 – the reaction of glucose fermentation occurs with the release of ethanol and carbon dioxide;
2. Determine M (C6H12O6) = 12 * 6 + 12 + 16 * 6 = 180 g / mol;
3. Let’s convert decimeters to liters, taking into account the tabular data:
1 dm3 = 1 l, which means 450 dm3 (CO2) = 450 l. (CO2);
4. Determine the amount of CO2 moles according to Avogadro’s law:
1 mol gas arr. in-va – 22.4 liters. at n. at.
X mol (CO2) -450 l. hence, X mol (CO2) = 1 * 450 / 22.4 = 20 mol;
5. Let’s make the proportion:
X mol (C6H12O6) – 20 mol (CO2);
– 1 mol – 2 mol from here, X mol (C6H12O6) = 1 * 20/2 = 10 mol;
6. Let’s calculate the mass of glucose by the formula:
M (C6H12O6) = Y * M = 10 * 180 = 1800 g.
Answer: 1800 g of glucose was required to carry out the reaction.