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Bacterial digestion is an economical method of sewage treatment. The reaction Bacteria SCO2(g) + 55NH4+ (aq) + 7602(g) CsH,ON(s) + 54NO2 (aq) + 52H2O(l) + 109H+ (aq) Bacterial tissue is an intermediate step in the conversion of the nitrogen in organic compounds to nitrate ions. How much bacterial tissue is produced in a treatment plant for every 1.0 x 104 kg of wastewater containing 3.0% NH4+ ions by mass? Assume that 95% of the ammonium ions are consumed by the bacteria.

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Final answer:

To determine how much bacterial tissue is produced in a treatment plant from 1.0 x 10^4 kg of wastewater containing 3.0% NH4+ ions, one must calculate the total mass of NH4+ consumed and then use stoichiometry to find the resulting mass of bacterial tissue. Approximately 32.5 kg of bacterial tissue is produced assuming 95% of NH4+ is consumed by the bacteria.

Step-by-step explanation:

The question asks us to calculate how much bacterial tissue is produced after processing wastewater containing ammonium ions, given that the bacteria consume 95% of the ammonium ions. Here's the step-by-step calculation:


  • First, identify the mass of NH4+ in the wastewater: 1.0 x 10^4 kg wastewater × 0.03 = 300 kg NH4+.

  • Then, calculate the mass of NH4+ consumed by bacteria: 300 kg × 0.95 = 285 kg.

  • From the balanced equation, 55 moles of NH4+ result in 1 mole of C5H7ON. Using the molar mass of NH4+ (18.04 g/mol), the mass of NH4+ per mole of C5H7ON is 55 × 18.04 g/mol × 1 kg/1000 g = 0.9912 kg.

  • Calculate the moles of C5H7ON produced: (285 kg / 0.9912 kg/mol) = 287.5 moles.

  • Finally, using the molar mass of C5H7ON (113 g/mol), the mass of bacterial tissue produced is: 287.5 moles × 113 g/mol × 1 kg/1000 g ≈ 32.5 kg.

Therefore, a treatment plant will produce approximately 32.5 kg of bacterial tissue for every 1.0 x 10^4 kg of wastewater treated, with the assumption that 95% of the ammonium ions are used by the bacteria.

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