How many g of glucose is burned if 4.4 g of CO2 is obtained.

univerkov | December 9, 2020 | education

C6H12O6 + 6O2 -> 6CO2 + 6H2O m (CO2) = 4.4g
m (C6H12O6) -?
n (CO2) = m / M = 4.4g / 44g / mol = 0.1 mol n (CO2) = n (C6H12O6) = 0.1 mol m (C6H12O6) = n * M = 0.1 mol * 180g / mol = 18g
Answer: 18g

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