Asked by Lex

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.

Answers

Answered by bobpursley
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.
Answered by Anonymous
.120

is that the answer?
Answered by Anonymous
No it's not
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