Bacterial digestion is an economical method of sewage treatment. The reaction

5CO2(g) + 55NH4+(aq) + 76O2(g) --bacteria--> C5H7O2N(s)[bacterial tissue] + 54NO2-(aq) + 52H2O(l) + 109H+(aq)

is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. How much bacterial tissue is produced in a treatment plant for every 1.0 x 10^4 kg of wastewater containing 3.0% NH4+ ions by mass? Assume that 95% of the ammonium ions are consumed by the bacteria.

From the balanced equation, you know that 55*.95moles of ammonium ions yields one mole of tissue.

Compute molmassNH4, and molemassBacterialTissue, from the formulas.

Bacterialtissue=10^4kg*.03(1000g/kg)*1moleammonium/molmassammonium*1moletissue/55*.95moleammonium *molemasstissue/1moletissue

Doesn't that do int in grams? Check my thinking.

.120

is that the answer?

No it's not

To find the amount of bacterial tissue produced, we need to calculate the number of moles of NH4+ ions present in the wastewater and then determine the moles of bacterial tissue produced using the balanced equation.

Step 1: Calculate the mass of NH4+ ions in the wastewater:
Mass of NH4+ ions = (3.0% x 1.0 x 10^4 kg) = 3.0 x (1.0 x 10^4 kg) / 100 = 3.0 x 10^2 kg

Step 2: Calculate the number of moles of NH4+ ions:
Molar mass of NH4+ = 14.01 g/mol + 4(1.01 g/mol) = 18.05 g/mol
Number of moles of NH4+ ions = Mass of NH4+ ions / molar mass of NH4+
Number of moles of NH4+ ions = (3.0 x 10^2 kg) / (18.05 g/mol) = (3.0 x 10^2 kg) / (0.01805 kg/mol) = 1.66 x 10^4 mol

Step 3: Calculate the number of moles of bacterial tissue produced:
From the balanced equation, we can see that for every 55 moles of NH4+ ions consumed, 1 mole of bacterial tissue is produced.
Number of moles of bacterial tissue = (1.66 x 10^4 mol) x (1 mol / 55 mol) = 3.02 x 10^2 mol

Step 4: Convert moles of bacterial tissue to mass:
Molar mass of bacterial tissue (C5H7O2N) = (12.01 g/mol x 5) + (1.01 g/mol x 7) + (16.00 g/mol x 2) + (14.01 g/mol) = 113.11 g/mol
Mass of bacterial tissue = Number of moles of bacterial tissue x molar mass of bacterial tissue
Mass of bacterial tissue = (3.02 x 10^2 mol) x (113.11 g/mol) = 3.42 x 10^4 g

Therefore, the amount of bacterial tissue produced in the treatment plant for every 1.0 x 10^4 kg of wastewater with 3.0% NH4+ ions is 3.42 x 10^4 grams.