Question

Для получения 270 г глюкозы (75% выход) необходима сахароза массой

Answer 1

Масса глюкозы теоретическа: 270/0,75=360г, т.е. 2 моль
     х г                               2 моль
С12Н22О11 +Н2О = С6Н12О6 + С6Н12О6 ( глюкоза и фруктоза)
  342 г                            1 моль
масса сахарози = 684г

Answer 2

Calculating Mass of Sucrose Required to Obtain 270g of Glucose (with 75% Yield)

To calculate the mass of sucrose required to obtain 270g of glucose with a 75% yield, we can use the information provided in the search results.

The formula for the mass of sucrose required can be calculated using the following steps:

1. Determine the Formula for Calculating the Mass of Sucrose: - The formula for calculating the mass of sucrose required to obtain a certain amount of glucose with a given yield is given by the equation: Mass of Sucrose = (Mass of Glucose) / (Yield of Glucose) * (Molar Mass of Sucrose / Molar Mass of Glucose)

2. Apply the Given Values: - Mass of Glucose = 270g - Yield of Glucose = 75%

3. Find the Molar Mass of Sucrose and Glucose: - The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol. - The molar mass of glucose (C6H12O6) is approximately 180.2 g/mol.

4. Calculate the Mass of Sucrose: - Using the given values and the formula, we can calculate the mass of sucrose required.

Applying the formula and the given values, the mass of sucrose required can be calculated as follows:

Mass of Sucrose = (270g) / (0.75) * (342.3 g/mol / 180.2 g/mol)

Mass of Sucrose = (270 / 0.75) * (342.3 / 180.2) ≈ 648g

Therefore, approximately 648 grams of sucrose are required to obtain 270 grams of glucose with a 75% yield.

This calculation is based on the assumption that the reaction proceeds with 100% efficiency, which may not always be the case in practical scenarios.