What volume of carbon monoxide will be released during alcoholic fermentation of glucose with an amount of 5 mol?
July 26, 2021 | education
| 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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