Question

Опредилите обьем углекислого газа каторый выделится при браженние 180 тонн глюкозы.

Answer 1

С6Н12О6 = 2С2Н5ОН + 2СО2
180
находим количество вещества С6Н12О6 = 180/180=1моль
находим количество вещества СО2= 2моль
находим объем СО2 = 2моль*22,4л/моль = 44,8л

Answer 2

Calculating the volume of carbon dioxide produced from fermenting 180 tons of glucose

When glucose is fermented, the main reaction that occurs is:

C6H12O6 → 2C2H5OH + 2CO2

This reaction shows that for every molecule of glucose fermented, two molecules of carbon dioxide are produced.

The molar mass of glucose is 180.16 g/mol. To calculate the number of moles of glucose present:

Mass of glucose = 180 tons = 180,000 kg Moles of glucose = Mass / Molar mass = 180,000 kg / 180.16 g/mol = 1000 moles

Since the reaction ratio is 2 moles of CO2 produced per 1 mole of glucose, the number of moles of CO2 produced will be:

Moles of CO2 = 2 * 1000 moles of glucose = 2000 moles

To calculate the volume of CO2 at standard temperature and pressure:

Volume of gas = Moles * Molar gas constant * Temperature / Pressure = 2000 moles * 0.0821 L*atm/mol*K * 273 K / 1 atm = 55,200,000 L = 55,200 m3

Therefore, the approximate volume of carbon dioxide produced from fermenting 180 tons of glucose is 55,200 cubic meters.