What volume of carbon monoxide will be released during alcoholic fermentation of glucose with an amount of 5 mol?

Let’s compose the reaction equation, find the quantitative ratios of substances.

C6H12O6 → 2С2Н5ОН + 2СО2 ↑.

For 1 mole of glucose, there are 2 moles of carbon dioxide. Substances are in quantitative ratios 1: 2. The amount of carbon dioxide will be 2 times more than the amount of glucose.

n (СО2) = 2n (C6H12O6) = 2 × 5 = 10 mol.

Let’s find the volume of hydrogen.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 10 mol × 22.4 L / mol = 224 L.

Answer: V = 224 liters.

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