180 g of glucose underwent alcoholic fermentation. the gas obtained in this case was

| January 28, 2021 | education

180 g of glucose underwent alcoholic fermentation. the gas obtained in this case was completely absorbed by 800 g of a 20% sodium hydroxide solution. The mass of the salt obtained.

Let’s write down the reaction schemes and arrange the coefficients:

1.C6H12O6 = 2C2H5OH + 2CO2

2.2NaOH + CO2 = Na2CO3 + H2O or NaOH + CO2 = NaHCO3

Let us determine the amount of glucose substance n (gl) = m (gl) / M (gl), where m (gl) and M (gl) are the mass and molar mass of glucose, respectively.

n (ch) = 180/180 = 1 mol.

From the reaction equation (item 1) it can be seen that from 1 mol of glucose 2 mol of carbon dioxide is formed, that is, n (CO2) = 2n (ch) = 2 mol.

Let’s calculate the amount of sodium hydroxide substance n (NaOH) to find out the ratio of the reagents.

m (NaOH) = m (solution) * w / 100%, where m (NaOH) is the mass of sodium hydroxide, m (solution) is the mass of the solution, w is the mass fraction of sodium hydroxide.

n (NaOH) = m (NaOH) / M (NaOH) = m (solution) * w / (M (NaOH) * 100%), where M (NaOH) is the molar mass of sodium hydroxide.

n (NaOH) = 800 * 20% / (40 * 100%) = 4 mol.

Since the ratio of the reagents n (CO2): n (NaOH) = 2: 4 = 1: 2, the process is described by the first reaction from item 2, and the mass of sodium carbonate m (Na2CO3) = n (Na2CO3) * M (Na2CO3) where

n (Na2CO3) is the amount of Na2CO3 substance, n (Na2CO3) = n (CO2), and M (Na2CO3) is the molar mass of sodium carbonate.

m (Na2CO3) = 2 * 105 = 210 g.



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