Find the mass of glucose that is required to obtain 400 ml of 70% ethanol solution, density = 0.868

| February 16, 2021 | education

Find the mass of glucose that is required to obtain 400 ml of 70% ethanol solution, density = 0.868, if the fraction of ethanol yield during fermentation is 60%?

Find the mass of the ethanol solution.

m = Vp.

m = 400 ml × 0.868 = 347.2 g.

W = m (substance): m (solution) × 100%,

hence m (substance) = (m (solution) × w): 100%.

m (substance) = (347.2 g × 70%): 100% = 243.04 g.

243.04 g – 100%,

X – 60%,

X = (243.04 × 60%): 100% = 145.82 g.

Let’s find the amount of substance С2Н5ОН.

M (C2H5OH) = 46 g / mol.

n = m: M.

n = 145.82: 46 g / mol = 3.17 mol.

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

С6Н12О6 → 2С2Н5ОН + 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.

C6H12O6.

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

n (C6H12O6) = ½ n (C2H5OH) = 3.17: 2 = 1.585 mol.

Let’s find the mass of С6Н12О6.

M (C6H12O6) = 180 g / mol.

m = n × M.

m = 180g / mol × 1.585 mol = 285.3 g.

Answer: 285.3 g.



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