Question

рассчитайте объем оксида углерода(4) приведенный к нормальным условиям,который выделится при спиртовом брожении 225г. глюкозы

Answer 1

С6H12O6 = 2C2H5OH + 2CO2

 

кол-во глюкозы = 225/180 = 1,25моль

кол-во угл газа = 2* кол-во глюкозы = 2,5 моль

объем = кол-во * молярный объем = 2,5 * 22,4 = 56л

Answer 2

Calculating the Volume of Carbon Dioxide

To calculate the volume of carbon dioxide (CO2) produced during the fermentation of 225g of glucose, we can use the stoichiometry of the reaction. The balanced chemical equation for the fermentation of glucose is:

C6H12O6 → 2C2H5OH + 2CO2

From this equation, we can see that for every mole of glucose fermented, 2 moles of carbon dioxide are produced.

Calculation

1. Calculate the moles of glucose: - The molar mass of glucose (C6H12O6) is 180.16 g/mol. - Therefore, the moles of glucose can be calculated as: ``` Moles of glucose = Mass of glucose / Molar mass of glucose = 225g / 180.16 g/mol ≈ 1.25 moles ```

2. Calculate the moles of carbon dioxide: - Since 2 moles of carbon dioxide are produced for every mole of glucose, the moles of carbon dioxide produced will be: ``` Moles of CO2 = 2 * Moles of glucose = 2 * 1.25 = 2.5 moles ```

3. Convert moles to volume at standard conditions: - At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. - Therefore, the volume of carbon dioxide produced at STP will be: ``` Volume of CO2 = Moles of CO2 * 22.4 L/mol = 2.5 * 22.4 = 56 liters ```

Conclusion

The volume of carbon dioxide produced when 225g of glucose is fermented at standard conditions is approximately 56 liters.