Calculate the mass of glucose that has undergone fermentation and the volume of carbon dioxide

Calculate the mass of glucose that has undergone fermentation and the volume of carbon dioxide obtained, if 526 g of ethanol were obtained.

Alcoholic fermentation of glucose can be written as an equation:

С6Н12О6 = 2 С2Н5ОН + 2 СО2

The molar masses of the substances are as follows:

glucose = 180 g / mol

ethanol = 46 g / mol

Molar volume of carbon dioxide = 22.4 l / mol

Using the reaction equation, we will compose the proportion to find the mass of glucose:

X g of glucose corresponds to 526 g of ethanol as

180 g / mol glucose corresponds to 2 * 46 g / mol ethanol

X = (526 * 180) / (2 * 46) = 1029 g of glucose

Now let’s calculate the volume of carbon dioxide.

526 g of ethanol corresponds to Y l. carbon dioxide as

2 * 46 g / mol corresponds to 2 * 22.4 l / mol of carbon dioxide

Y = (526 * 2 * 22.4) / (2 * 46) = 256 HP carbon dioxide.

Answer: 1029 g of glucose and 256 liters. carbon dioxide.