A lake with a cross-sectional area of 1 km 2 and a depth of 50 meters has a euph

Question

A lake with a cross-sectional area of 1 km2 and a depth of 50 meters has a euphotic zone that extends 15 meters below the surface. What is the maximum weight for the biomass (in grams of carbon) that can be decomposed by aerobic bacteria in the water column of the lake below the euphotic zone during the summer when there is no circulation in the upper layer? The bacterial decomposition reaction is:

{CH2O} + O2 CO2 + H2O

The solubility of oxygen in pure water saturated with air at 20 °C is 8.9 mg/L; 1 m3 = 1,000 liters.

Explanation / Answer

Cross-sectional area of lake = 1 km2 = 106 m2

Depth of euphotic zone = 15 m

Volume of euphotic zone in lake = 15x106 m3

= 1.5x1010 L

Solubility of oxygen in pure water saturated with air at 20 °C = 8.9 mg/L

Mass of oxygen dissoled in water = 8.9 * 1.5x1010 mg

= 1.335x1011 mg

= 1.335x108 g

Moles of oxygen dissolved = 1.335x108 g / 32 g/mole

= 4.171x106 moles

The bacterial decomposition reaction is:

{CH2O} + O2 CO2 + H2O

Since, 1 mole biomass is decomposed by 1 mole oxygen.

Moles of biomass decomposed by 4.171x106 moles oxygen = 4.171x106 moles

Atomic mass of C = 12 g/mole

Mass of biomass decomposed (in grams of carbon) = 4.171x106 * 12

= 5.005x107 g

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.