Citric acid in industry is obtained by microbiological fermentation of a glucose solution according to the equation

Citric acid in industry is obtained by microbiological fermentation of a glucose solution according to the equation: 2C6H12O6 + 3O2 – 2C6H8O7 + 4H2O How many kg of citric acid with a yield of 62% of the theoretically possible can be obtained from 520kg of 15% glucose solution?

Find the mass of glucose in the solution.

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

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

m (substance) = (520 kg × 15%): 100% = 78 kg.

Let’s find the amount of substance С6Н12О6.

n = m: M.

M (C6H12O6) = 180 kg / kmol.

n = 78 kg: 180 kg / kmol = 0.43 kmol.

Let’s find the quantitative ratios of substances.

2C6H12O6 + 3O2 → 2C6H8O7 + 4H2O.

For 2 kmol of С6Н12О6, there is 2 kmol of С6Н8О7.

Substances are in quantitative ratios 1: 1.

The amount of substances will be the same.

n (C6H12O6) = n (C6H8O7) = 0.43 kmol.

Let’s find the mass of С6Н8О7.

M (C6H8O7) = 192 kg / kmol.

m = n × M.

m = 192 kg / kmol × 0.43 kmol = 82.56 kg.

82.56 g of C6H8O7 were obtained according to calculations. According to the condition of the problem, the yield was 62%. Let’s find the mass of С6Н8О7 from the theoretically possible.

82.56 – 100%.

m – 62%.

m = (62% × 82.56%): 100% = 51.19 kg.

Answer: 51.19 kg.