The emission of N O 2 by fossil fuel combustion can be prevented by injecting ga

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 2.42 L/s at 644 K and contains a partial pressure of NO of 12.7 torr .

Part A

What total mass of urea is necessary to react completely with the NO formed during 7.3 hours of driving? Express your answer using two significant figures. 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 2.42 L/s at 644 K and contains a partial pressure of NO of 12.7 torr . 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 2.42 L/s at 644 K and contains a partial pressure of NO of 12.7 torr . 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 2.42 L/s at 644 K and contains a partial pressure of NO of 12.7 torr . 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 2.42 L/s at 644 K and contains a partial pressure of NO of 12.7 torr .

Part A

What total mass of urea is necessary to react completely with the NO formed during 7.3 hours of driving? Express your answer using two significant figures.

Part A

What total mass of urea is necessary to react completely with the NO formed during 7.3 hours of driving? Express your answer using two significant figures.

Part A

What total mass of urea is necessary to react completely with the NO formed during 7.3 hours of driving? Express your answer using two significant figures.

Part A

What total mass of urea is necessary to react completely with the NO formed during 7.3 hours of driving? Express your answer using two significant figures.

Part A

What total mass of urea is necessary to react completely with the NO formed during 7.3 hours of driving? Express your answer using two significant figures.

Part A

What total mass of urea is necessary to react completely with the NO formed during 7.3 hours of driving? Express your answer using two significant figures. What total mass of urea is necessary to react completely with the NO formed during 7.3 hours of driving? Express your answer using two significant figures. 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 2.42 L/s at 644 K and contains a partial pressure of NO of 12.7 torr .

Part A

What total mass of urea is necessary to react completely with the NO formed during 7.3 hours of driving? Express your answer using two significant figures.

Explanation / Answer

Driving time= 7.3 hours, converting this into seconds, 60 min= 1hr, 1 min= 60 sec

7.3 hrs= 7.3hr*60min/hr* 60 sec/min=26280 seconds

flow rate of exhaust= 2.42 L/s, volme of Exhaust= flow rate of exhaust* time in seconds= 2.42*26280 L=63598 Liters

This is V= 63598 L, P= partial pressure which is defined as the pressure exerted if it alone occupies the entire volume= 12.7 Torr, 760 Torr= 1atm, 12.7 Torr= 12.7/760 atm=0.0167 atm, T= 644K, R= gas constant =0.0821 L.atm/mole.K, n= no of moles = PV/RT= 0.0167*63598/(0.0821*644) =20.1 moles of NO is there in the exhaust.

From the reaction given, 4 moles of NO requires 2 mole of Urea

20.1 moles of NO requires 20.1*2/4= 10.05 moles of Urea

molar mass of Urea (NH2CO.NH2)= 60 g/mole, mass of urea= moles* molar mass=10.05*60 gm =603 gm

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