Question

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)+2CO2(g)+4H2O(g)
Suppose that the exhaust stream of an automobile has a flow rate of 2.56 L/s at 672 K and contains a partial pressure of NO of 13.5 torr .

a.) What total mass of urea is necessary to react completely with the NO formed during 8.2 hours of driving?

Answer 1

Flow rate = 2.56 L/s

Driving time = 8.2 h * (60 min/ 1 h) * (60 s / 1 min)

                 = 29520 s

So volume of exhaust = 2.56 L/s * 29520 s

                              = 75571. 2 L

Assuming total pressure of exhaust = 1.0 atm,

Moles of exhaust gas,

n = PV / RT

= 1.00 atm * 75571. 2 L / (0.0821 L-atm/mol.K * 672 K)

   = 1369. 78 moles

Partial pressure of NO = 13.5 torr

Partial pressure of NO = Total pressure * mole fraction

13.5 torr = 760 torr * mf

So mole fraction of NO = 0.0149

Mole fraction of NO = moles of NO / total moles

0.0149 = moles of NO / 1369.78 moles

So moles of NO = 20.409 moles

Mass of urea required

= 20.409 moles NO * (2 moles urea/ 4 moles NIO) * (60.06 g / 1 mole urea)

= 612. 88 g ≈ 613 g