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

ID: 504034 • Letter: T

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? 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? 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?

Explanation / Answer

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 + 10SO42-+ 5Fe2+ + 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.

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