Question

At sea-level and at 30.0 °C, the solubility of O₂ in water is 7.50 mg/L. Consider a water body at that temperature containing 6.05 mg/L of oxygen. By photosynthesis, 1.14 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.

Answer 1

The reaction is:

6 CO₂ + 6 H₂O = C₆H₁₂O₆ + 6 O₂

Moles of O2 produced = moles of CO2 reacted

Moles of CO2 present per liter = 1.14 mg/ Molar mass of CO2 = 1.14 mg/ (44 mg/mili mole) = 0.026 milimoles

Moles of O2 produced in one liter = 0.026 moles

Mass of O2 in one liter = 0.026 * molar mass of O2 = 0.026 mili mole * 32 mg /mili mole = 0.83 mg

SO, total amount of O2 = 6.50 mg/L + 0.83 mg/ L = 7.33 mg/ L

which does not exceed the solubility of O₂ in water (7.50 mg/L)