Question

При спиртовом брожении глюкозы выделилось 5.6 л углекислого газа. Чему равна масса глюкозы,подвергшейся брожению?

Answer 1

       X                               5.6
C6H12O6=2C2H5OH+2CO2
      180                           44.8

X=180*5.6/44.8=180/8=22.5 грамм CO2

Answer 2

Calculating the Mass of Glucose Subjected to Fermentation

To calculate the mass of glucose that underwent fermentation, we can use the ideal gas law to relate the volume of carbon dioxide produced to the amount of glucose consumed.

The balanced chemical equation for the fermentation of glucose is: C6H12O6 → 2C2H5OH + 2CO2

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

Given that 5.6 L of carbon dioxide was produced, we can calculate the amount of glucose consumed.

Using the ideal gas law equation: PV = nRT Where: - P = pressure - V = volume - n = number of moles - R = ideal gas constant - T = temperature

We can rearrange the equation to solve for the number of moles: n = PV / RT

Assuming standard temperature and pressure (STP), where T = 273 K and P = 1 atm, and using the value of R = 0.0821 L·atm/(K·mol), we can calculate the number of moles of carbon dioxide produced.

Given that 1 mole of glucose produces 2 moles of carbon dioxide, we can then calculate the mass of glucose consumed.

Calculation

Using the ideal gas law equation: n = PV / RT

Given: - V = 5.6 L - P = 1 atm - T = 273 K - R = 0.0821 L·atm/(K·mol)

We can calculate the number of moles of carbon dioxide produced: n = (5.6 L) * (1 atm) / (0.0821 L·atm/(K·mol) * 273 K)

n = 0.22 moles

Since 1 mole of glucose produces 2 moles of carbon dioxide, the moles of glucose consumed is: 0.22 moles * (1 mole glucose / 2 moles CO2) = 0.11 moles

The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol.

Therefore, the mass of glucose consumed is: 0.11 moles * 180.16 g/mol = 19.82 grams

So, the mass of glucose that underwent fermentation is approximately 19.82 grams.