What mass of ethanol is formed during alcoholic fermentation of glucose, weighing 250 g, containing 4% impurities.

Let’s find the mass of pure glucose without impurities.

100% – 4 5 = 96%.

250 g – 100%,

X – 96%.

X = (250 × 96%): 100% = 240 g.

Let’s find the amount of substance С6Н12О6.

n = m: M.

M (C6H12O6) = 180 g / mol.

n = 240 g: 180 g / mol = 1.33 mol.

Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.

С6Н12О6 → 2 С2Н5ОН + 2СО2 ↑.

According to the reaction equation, there is 2 mol of C2H5OH per 1 mol of С6Н12О6. The substances are in quantitative ratios of 1: 2.

The amount of C2H5OH substance is 2 times more than C6H12O6.

n (C2H5OH) = 2n (C6H12O6) = 1.33 × 2 = 2.66 mol.

Let’s find the mass of С2Н5ОН.

M (C2H5OH) = 46 g / mol.

m = n × M.

m = 46 g / mol × 2.66 mol = 122.36 g.

Answer: 122.36 g.