Bacterial digestion is an economical method of sewage treatment. in the reaction, given below, bacterial tissue is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. what mass of 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. 5co2(g) + 55nh4+(aq) + 76o2(g) c5n7o2n(s) + 54no2-(aq) + 52h2o(l) + 109h+(aq)

1. Solve the mass of NH4+ ions = (3%/100%)(1x10^4kg) = 300 kg = 300,000 g 

2. Compute for the amount that will be converted as a bacterial tissue: 
NH4+ ions consumed by bacteria = (95%/100%)(300,000 g) = 285,000 g 

3. Convert the mass of NH4+ ion calculated in step 2 into the # of moles: 
moles of NH4+ = mass of NH4+ / MW of NH4 

moles of NH4+ = 285,000 g / 18 g/mol = 15,833.33 mol
the stoichiometric ratio of C5H7O2N and NH4 is 1 : 55 grounded on the given balanced chemical equation: 
moles of C5H7O2N = moles of NH4+ x (1 mole of C5H7O2N / 55 moles of NH4+) 

moles of C5H7O2N = 15,833.33 moles x (1/55) = 303.03 moles 

4. Convert the moles of C5H7O2N into grams: 
mass of C5H7O2N = # of moles of C5H7O2N x MW of C5H7O2N 

= 303.03 moles x 113 g/mol 

= 34,242.39 grams

= 34.34 kg 

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Sorry i dont understand the question can you rephase it?