Answer to Question #200713 in General Chemistry for lea

Question #200713

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 2.32 L/s at 658 K and contains a partial pressure of NO of 10.6 torr .


What total mass of urea is necessary to react completely with the NO formed during 8.4 hours of driving? Express your answer using two significant figures.


1
Expert's answer
2021-05-31T02:05:33-0400

8.4 hours × 3600 seconds / hour = 30240 seconds

30240 seconds × 2.32 L/s = 70156.8 Litres of NO

Find the moles of NO using the ideal gas law:

PV = nRT

(10.6Torr) × (70156.8 Litres ) = n × (62.36 L∙Torr/mol∙K) × (658K)

n = 18.12 moles of NO

According to the equation

2CO(NH2)2(g) + 4NO(g) + O2(g) → 4N2(g) + 2CO2(g)+4H2O(g)

4 moles of NO reacts with 2 moles of urea,

which, using molar mass, is

2 mol urea × 60.06 g/mol = 120.12 grams

so

18.12 moles of NO × (120.12 grams urea) / (4 mol NO) = 544.14 grams of urea


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