Bacterial digestion is an economical method of sewage treatment.
5CO2(g) + 55NH4+(aq) + 76O2(g) -> C5H7O2N(s) + 54NO2-(aq) + 52H2O(l) + 109H+(aq)
bacterial tissue
The above reaction is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. a. How much bacterial tissue is produced in a treatment plant for every 1.0 x 10^4 of wastewater containing 3.0% NH4+ ions by mass? b. Assume that 95% of the ammonium ions are consumed by the bacteria.
To calculate the amount of bacterial tissue produced, we need to use stoichiometry. The balanced equation shows that 55 moles of NH4+ ions react with 5 moles of CO2 and 76 moles of O2 to produce 1 mole of bacterial tissue (C5H7O2N), among other products.
a. To find the amount of bacterial tissue produced for 1.0 x 10^4 g of wastewater containing 3.0% NH4+ ions, we can start with the mass of NH4+ ions in the wastewater and convert it to moles.
Mass of NH4+ ions = 1.0 x 10^4 g x 0.03 = 300 g
Molar mass of NH4+ = 14.01 g/mol + 1.01 g/mol = 15.02 g/mol
Moles of NH4+ ions = Mass of NH4+ ions / Molar mass of NH4+ = 300 g / 15.02 g/mol = 19.98 mol
Now we can use the stoichiometry of the balanced equation to find the moles of bacterial tissue produced.
Moles of bacterial tissue = (19.98 mol NH4+) x (1 mol bacterial tissue / 55 mol NH4+) = 0.363 mol
b. Assuming 95% of the ammonium ions are consumed by the bacteria, we can multiply the moles of bacterial tissue calculated in part a by the consumption efficiency.
Moles of bacterial tissue produced = 0.363 mol x 0.95 = 0.345 mol
Therefore, for every 1.0 x 10^4 g of wastewater containing 3.0% NH4+ ions, approximately 0.345 mol of bacterial tissue is produced.