Question

В результате брожения глюкозы массой 90 г образовался этанол массой 36.8 г. Определите практический выход этанола

Answer 1

C6H12O6----ферменты-->2C2H5OH+2H2O

Дано:
m(C6H12O6)=90гр
m(C2H5OH)=38,8гр
Найти n(этта)
Практической выход
Решение:
n(C6H12O6))=m/M=90/180=0,5моль
По уравнению реакции спирта в 2 раза больше, т.е на 1моль глюкозы .2моль спирта:
n(C2H5OH)=0,5*2=1моль
m(C2H5OH)=n*M=1*46г/моль=46гр
Но в условии сказано , что практической выход спирта36,8гр
n(этта)=m(практическая)/m(теоретическую)*100=
36,8/46*100=80%
Ответ: выход этанола 80% от теоретически возможного

Answer 2

Calculation of Ethanol Yield

To determine the practical yield of ethanol, we need to compare the mass of ethanol produced to the theoretical maximum yield based on the amount of glucose used.

Given: - Mass of glucose = 90 g - Mass of ethanol produced = 36.8 g

To calculate the practical yield of ethanol, we can use the following formula:

Practical Yield (%) = (Mass of ethanol produced / Theoretical maximum yield) x 100

The theoretical maximum yield of ethanol can be calculated based on the stoichiometry of the reaction. The balanced equation for the fermentation of glucose to ethanol is:

C6H12O6 → 2C2H5OH + 2CO2

From the equation, we can see that for every mole of glucose, we should obtain 2 moles of ethanol. To calculate the theoretical maximum yield, we need to convert the mass of glucose to moles and then use the mole ratio to determine the moles of ethanol produced.

Calculation Steps:

1. Calculate the number of moles of glucose: - The molar mass of glucose (C6H12O6) is 180.16 g/mol. - Moles of glucose = Mass of glucose / Molar mass of glucose

2. Calculate the theoretical moles of ethanol produced: - The mole ratio between glucose and ethanol is 1:2. - Moles of ethanol = Moles of glucose x 2

3. Calculate the theoretical mass of ethanol produced: - Mass of ethanol = Moles of ethanol x Molar mass of ethanol

4. Calculate the practical yield of ethanol: - Practical Yield (%) = (Mass of ethanol produced / Theoretical mass of ethanol) x 100

Let's perform the calculations:

1. Moles of glucose = 90 g / 180.16 g/mol = 0.4997 mol (rounded to four decimal places)

2. Moles of ethanol = 0.4997 mol x 2 = 0.9994 mol (rounded to four decimal places)

3. The molar mass of ethanol (C2H5OH) is 46.07 g/mol. Mass of ethanol = 0.9994 mol x 46.07 g/mol = 45.99 g (rounded to two decimal places)

4. Practical Yield (%) = (36.8 g / 45.99 g) x 100 = 80.0% (rounded to one decimal place)

Therefore, the practical yield of ethanol in this reaction is approximately 80.0%.