Calculate the mass of air required to completely oxidize 90 g of glucose.

univerkov | April 8, 2021 | education

Given:
m (C6H12O6) = 90 g

To find:
m (air) -?

Decision:
1) C6H12O6 + 6O2 => 6CO2 + 6H2O;
2) n (C6H12O6) = m / M = 90/180 = 0.5 mol;
3) n (O2) = n (C6H12O6) * 6 = 0.5 * 6 = 3 mol;
4) m (O2) = n * M = 3 * 32 = 96 g;
5) m (air) = m (O2) * 100% / ω (O2) = 96 * 100% / 23% = 417 g.

Answer: The air mass is 417 g.

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