DanHalen
Well-known member
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 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