NCEES 2008 Problem 512 WR/Env

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DanHalen

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The problem states:

The fully nitrified influent to the anoxic basin contains 25 mg/L nitrate-N. The following equation applies:

NO3-+1.08CH3OH+H+-->0.065C5H7O2N+0.47N2+0.76CO2+2.44H2O

The minimum ethanol requirement (mg/L) to provide complete denitrification is most nearly:

A) 14

B) 23

C) 57

D) 62

SOLUTION

According to the stoichiometric equation, 1.08 moles of methanol are consumed by 1 mole of NO3-. The molecular weight of methanol is 32 and that of nitrate-nitrogen is 14 (Where exactly is the 14 coming from? The problem statement or the equation?)

Therefore the minimum amout of methanol required = (1.08*32)/(14)*25 mg/L = 61.71 mg/L, ANS. D

 
The NO3- ion contains one N atom (atomic weight 14). So, one ion of NO3- is equivalent to 14 g/mol of NO3- N. I am not perfectly happy with the awkward way I worded that, but that's where the 14 comes from

 
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