Calculate the mass of sucrose that underwent hydrolysis, taking into account the fact that lactic acid weighing 54 g

| September 17, 2021 | education

Calculate the mass of sucrose that underwent hydrolysis, taking into account the fact that lactic acid weighing 54 g was obtained from the formed glucose with a mass fraction of 60% yield

To solve this problem, we need the equations for the hydrolysis of sucrose [1] and the equation for the reaction of lactic acid fermentation of glucose [2].

[1] C12H22O11 + H2O = C6H12O6 (glucose) + C6H12O6 (fructose)

[2] C6H12O6 = 2 CH3-CHOH-COOH

First, we find how much lactic acid would be formed theoretically at 100% yield.

X = 100% * 54g / 60% = 90g.

Now, knowing the total yield of lactic acid after the fermentation reaction, by equation [2] we will find the mass of glucose. Molar mass of glucose = 180.16 g / mol, molar mass of lactic acid = 90.08 g / mol

180.16 g / mol of glucose corresponds to 2 * 90.08 g / mol of lactic acid as

Y g of glucose corresponds to 90 g of lactic acid

Y = (90 * 180.16) / (2 * 90.08) = 90 g.

We found that 90 g of glucose was involved in the fermentation reaction. It is this mass of glucose that is formed as a result of sucrose hydrolysis. Using the equation [1], we find the mass of sucrose. Molar mass of sucrose = 342.3 g / mol

342.3 g / mol of sucrose corresponds to 180.16 g / mol of glucose as

Z g of sucrose corresponds to 90 g of glucose

Z = (90 * 342.3) /180.16 = 171 g.

Answer: 171 g of sucrose



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