The emission of no2 by fossil fuel combustion can be prevented by injecting gaseous urea into the combustion mixture. the urea reduces no (which oxidizes in air to form no2) according to the following reaction: 2co(nh2)2(g)+4no(g)+o2(g)→4n2(g)+2c o2(g)+4h2o(g) suppose that the exhaust stream of an automobile has a flow rate of 2.32 l/s at 670 k and contains a partial pressure of no of 12.1 torr. what total mass (in gram) of urea is necessary to react completely with the no formed during 8.1 hours of driving?

The mass of urea that is necessary to react completely is 587.68 g.

Explanation:

We are given:

Rate of flow of exhaust stream = 2.32 L/s

Total time of driving = 8.1 hrs

Converting this into seconds:

Total time of driving =

If, in 1 second, the volume of stream flows is 2.32 L

The, in 29160 seconds, the volume of stream flows will be =

Let us assume that the total pressure is 1 atm.

P = pressure = 1 atm

V = volume = 67651.2 L

n = number of moles = ?

R = Gas constant =

T = temperature = 670 K

Putting values in above equation, we get:

= partial pressure of NO = 12.1 torr

= mole fraction of NO = ?

= total pressure of solution = 1 atm = 760 torr   (Conversion factor: 1 atm = 760 torr)

Putting values in above equation, we get:

Mole fraction of NO = 0.0159

Total moles = 1231.6 moles

Putting values in above equation, we get:

By Stoichiometry of the reaction:

4 moles of nitrogen oxide reacts with 2 moles of urea.

So, 19.57 moles of nitrogen oxide will react with = of urea.

Moles of urea = 9.785 mol

Molar mass of urea = 60.06 g/mol

Putting values in above equation, we get:

Hence, the mass of urea that is necessary to react completely is 587.68 g.

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