How much carbon dioxide will be released during fermentation of 300 g of starch containing 90% glucose?

univerkov | August 9, 2021 | education

Given:
m (starch) = 300 g
ω (C6H12O6) = 90%

To find:
V (CO2) -?

1) C6H12O6 => 2C2H5OH + 2CO2 ↑;
2) m (C6H12O6) = ω (C6H12O6) * m (starch) / 100% = 90% * 300/100% = 270 g;
3) n (C6H12O6) = m / M = 270/180 = 1.5 mol;
4) n (CO2) = n (C6H12O6) * 2 = 1.5 * 2 = 3 mol;
5) V (CO2) = n * Vm = 3 * 22.4 = 67.2 liters.

Answer: The CO2 volume is 67.2 liters.

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