0.4 moles of average photosynthetic biomass falls into a 1000 L volume of Pacifi

Question

0.4 moles of average photosynthetic biomass falls into a 1000 L volume of Pacific Deep Water with an initial dissolved oxygen content of 150 muM. Assuming "average" phytoplankton grow with the stoichiometry of: 106CO2+ 16HNO3+ 1H3PO4+ 122H2O rightarrow CH2O)106(NH3)l6(H3PO4)l + 13802 As bacteria start to oxidize this tissue, which reactant will run out first, oxygen or organic tissue? From the stoichiometry of average marine organic tissue, calculate the increase in the concentrations of nitrate, phosphate, and total dissolved inorganic carbon you would expect from this oxidation.

Explanation / Answer

106CO2 + 16HNO3 + H3PO4 + 122H2O --------> (CH2O)106(NH3)16(H3PO4) + 138O2

moles of dissolved oxygen = molarity*volume of solution in litres = = (150*10-6)*1000 = 0.15 moles

moles of biomass = 0.4 moles

Thus, as per the balanced reaction ; biomass and O2 react in the molar ratio of 1:138

Clearly, for 0.4 moles of biomass, moles of O2 required = 0.4*138 = 55.2

Thus, O2 is limiting ragent and hence will run ou first.

Now, nitrate produced = (16/138)*moles of O2 reacted = (16/138)*0.15 = 0.0174

moles of H3PO4 produced = (1/138) moles of O2 consumed = 0.15/138 = 0.0011

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