As a result of alcoholic fermentation of glucose, 100 g of ethanol was obtained, the yield of which was 92%.

| February 11, 2021 | education

As a result of alcoholic fermentation of glucose, 100 g of ethanol was obtained, the yield of which was 92%. Enter the mass of glucose.

The alcoholic fermentation reaction of glucose can be written as follows:

С6Н12О6 = 2 С2Н5ОН + 2 СО2

Those. in one mole of glucose, 2 moles of ethanol are obtained. The molar mass of glucose is 180.16 g / mol, ethanol is 46.07 g / mol.

At the exit, we received 100 g of alcohol with a practical yield of 92%. We find the theoretical alcohol yield: 100 g * 100% / 92% = 108.7 g.

Knowing the molar masses of substances and the theoretical mass of the reaction product, we find the mass of glucose for fermentation according to the proportion:

180.16 g / mol of glucose corresponds to 2 * 46.07 g / mol of ethanol as

X g of glucose corresponds to 108.7 g of ethanol

X = (108.7 * 180.16) / (2 * 46.07) = 212.54 g.

Answer: 212.54 g glucose



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