3.0% NH4^+ ion 1.0×10^4 kg wastewater 95%

Bacterial digestion is an economical method of sewage treatment.

The reaction 5CO2(g) + 55NH4^+(aq) + 76O2(g) →

C5H7O2N(s) + 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 1.0×10^4 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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폐수 속 NH4^+ 이온의 질량을 계산하면,

(1.0×10^4 kg) × (3.0/100)

= 300 kg

박테리아에 의해 처리되는 NH4^+ 이온의 질량을 계산하면,

300 kg × (95/100)

= 285 kg

5CO2(g) + 55NH4^+(aq) + 76O2(g) →

C5H7O2N(s) + 54NO2^-(aq) + 52H2O(l) + 109H^+(aq)

NH4^+ : C5H7O2N = 55 : 1 계수비이므로,

질량비로 박테리아 조직의 질량을 계산하면,

( 참고: 질량비로 계산 https://ywpop.tistory.com/5078 )

> NH4^+의 몰질량 = 18.04 g/mol

> C5H7O2N의 몰질량 = 113.1146 g/mol

55NH4^+ : C5H7O2N = 55(18.04) : 113.1146 = 285 kg : ? kg

? = 113.1146 × 285 / (55(18.04))

= 32.49 kg

답: 32.5 kg

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