How much ethanol can be obtained from 42.5 kg of glucose, which contains 15% impurities?

Given:
m (C6H12O6) = 42.5 kg = 42500 g
ω approx. = 15%

Find:
m (C2H5OH) -?

Solution:
1) C6H12O6 => 2C2H5OH + 2CO2;
2) ω (C6H12O6) = 100% – ω approx. = 100% – 15% = 85%;
3) m clean. (C6H12O6) = ω (C6H12O6) * m tech. (C6H12O6) / 100% = 85% * 42500/100% = 36125 g;
4) n (C6H12O6) = m pure. (C6H12O6) / M (C6H12O6) = 36125/180 = 200.69 mol;
5) n (C2H5OH) = n (C6H12O6) * 2 = 200.69 * 2 = 401.38 mol;
6) m (C2H5OH) = n (C2H5OH) * M (C2H5OH) = 401.38 * 46 = 18463.5 g.

Answer: The mass of C2H5OH is 18463.5 g.