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