Question

At sea-level and at 30.0 °C, the solubility of oxygen, O2, in water is 7.50 mg/L. Consider a water body at that temperature containing 6.96 mg/L of oxygen. By photosynthesis, 1.96 mg/L of CO2 is converted to organic biomass, which has a composition of C6H12O6, during a single hot day. You want to assess whether the amount of oxygen produced is sufficient to exceed its solubility. Determine the amount of oxygen produced by photosynthesis on that day (in mg/L). (after Hites, Chaper 5, Problem 19; note that it should read '... 1.5 mg/L of CO2 ...' in the original Hites problem.)

Answer 1

The reaction is:

6 CO2 + 6 H2O = C6H12O6 + 6O2

From balanced stoichiometry; we get:

Moles of CO2 : Moles of O2 =6:6

So; moles of O2 = moles of CO2

Amount of CO2 = 1.96 mg/ L

So number of moles of CO2 = mass per liter/ molar mass of CO2

So; moles if CO2 = (1.96 mg/ Liter) / ( 44 mg/ mmol) = 0.0445 mmol

So; moles of O2 produced per liter = 0.0445 mmol/ liter

Mass of O2 produced during photosynthesis :

0.0445 mmol× 32 mg/ mmol = 1.42 mg/ Liter

So total mass of O2 per liter = 1.42 + 6.96 mg/ liter = 8.38 mg/ liter which exceeds the solubility (7.50 mg/ L)