Question

During mid-January 2004, average gas concentrations measured in Cache Valley were:

ammonium (NH4) = 16 µg/m^3

nitrate (NO3) = 45 µg/m^3

NH4 + NO3 = NH4NO3 (s)

a) Is NH4 or NO3 the limiting factor for the formation of NH4NO3? Explain using appropriate calculations.

b) Calculate the average NH4NO3 concentration (µg/m^3) based on these concentrations.

c) If we could reduce ammonium emissions by 15% by changing the composition of cattle feed, but nitrate remained the same, calculate the average NH4NO3 concentration (µg/m^3).

d) If we could reduce nitrate emissions by 15% through a vehicle inspection program, but ammonia remained the same, calculate the average NH4NO3 concentration (µg/m^3).

e) Based on your answers to part c) and d), which pollutant (NH4 or NO3) should we target first?

PS. This is an environmental engineering.

Answer 1

ammonium (NH4) = 16 µg/m^3

nitrate (NO3) = 45 µg/m^3

NH4 + NO3 = NH4NO3 (s)

a) Is NH4 or NO3 the limiting factor for the formation of NH4NO3? Explain using appropriate calculations.

Solution :-

Lets calculate the moles of the each reactant

Using the given concentrations

Moles of NH4 = (16 ug * 1 g / 1*10^6 ug)*(1 mol / 18.04 g ) = 8.87*10^-7 mol NH4

Moles of NO3 = (45 ug * 1 g / 1*10^6 ug)*(1 mol / 62.05 g)= 7.25*10^-7 mol NO3

Mole of the NO3 are less than moles of the NH4

Therefore NO3 is the limiting reactant since mole ratio is 1 :1

b) Calculate the average NH4NO3 concentration (µg/m^3) based on these concentrations.

Solution :- since the NO3 is the limiting reactant therefore using the mole ratio lets calculate the mass of the NH4NO3

(7.25*10^-7 mol NO3*1 mol NH4NO3/1 mol NO3)*(80.0434 g/ 1 mol NH4NO3) = 5.80*10^-5 g

5.80*10^-5 g * 1*10^6 ug / 1 g = 58 ug/m3

So the average concentration of the NH4NO3 = 58 ug/m3

c) If we could reduce ammonium emissions by 15% by changing the composition of cattle feed, but nitrate remained the same, calculate the average NH4NO3 concentration (µg/m^3).

Solution :-

If the concentration of the NH4 is decreased by 15 % then remaining is 100 – 15 = 85 %

So lets calculate the moles of the NH4NO3 that can be formed

(8.87*10^-7 mol NH4 * 1 mol NH4NO3/ 1 mol NH4)*(85 % / 100 %) = 7.54*10^-7 mol

Still the moles of the NH4 are more than moles of the NO3

Therefore the NO3 is the limiting reactant hence it will have the same concentration of the NH4NO3 as calculated in the part b that is

Concentration of the NH4NO3 = 58 ug/m3

d) If we could reduce nitrate emissions by 15% through a vehicle inspection program, but ammonia remained the same, calculate the average NH4NO3 concentration (µg/m^3).

Solution :-

NO3 is limiting reactant and if it is further decreased by the 15 % then amount of the formation of the NH4NO3 will decrease by 15 %

So the concentration of the Nh4NO3 that would be present is calculated as follows

58 ug * 85 % / 100 % = 49.3 ug/m3

Therefore the concentration of the NH4NO3 = 49.3 ug/m3

e) Based on your answers to part c) and d), which pollutant (NH4 or NO3) should we target first?

Solution :-

We need to target the NO3 reactant because decreasing its amount decreases the formation of the NH4NO3