Question

To fully denitrify 1000 gallons of water with a nitrate-N concentration of 17 mg N l-1 ...

a) how many grams of glucose would it take?

b) how many moles of methanol would be required?

c) how many moles of sulfide would be required?

Answer 1

Denitrification is a process where nitrates accept electrons i.e undergo reduction to produce molecular nitrogen or N₂. Denitrification is basically a redox process where nitrate will undergo reduction and the other entity in the reaction (such as sulfide in our case) will undergo oxidation (loss of electrons). Writing down a balanced reaction of nitrate reacting will glucose, methanol and sulfide helps in calculating the number of moles or grams as it gives us an idea of the stoichiometry of reactants.

a.) 5C₆H₁₂O₆ + 24NO₃^- = 24HCO₃^- + 6CO₂ + 12N₂ + 18H₂O

(1000 gallons X 3.78541 L/gallon) X (0.017 g of NO₃^- - N / 14 g per mol NO₃^- - N) X (1 mol of KNO₃ / 1 mol NO₃^- - N) X (5 mol C₆H₁₂O₆ / 24 mol NO₃^-) X (180.155 g of C₆H₁₂O₆ / 1 mol of C₆H₁₂O₆) = 172.5 g of glucose.

b.) 6NO₃^- + 5CH₃OH + 6H⁺ = 3N₂ + 5CO₂ + 13H₂O

(1000 gallons X 3.78541 L/gallon) X (0.017 g of NO₃^- - N / 14 g per mol NO₃^- - N) X (5 mol of CH₃OH / 6 mol of NO₃^-) = 3.83 moles of methanol.

c.) 14NO₃^- + 5FeS₂ + 4H⁺ = 7N₂ + 10SO₄^2-+ 5Fe^₂+ + 2H₂O

(1000 gallons X 3.78541 L/gallon) X (0.017 g of NO₃^- - N / 14 g per mol NO₃^- - N) X (5 mol of FeS₂ / 14 mol of NO₃^-) = 1.641 moles of sulfide.