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 2.37 L/s at 657 K and contains a partial pressure of NO of 14.0 torr. What total mass of urea is necessary to react completely with the NO formed during 8.1 hours of driving? Express answer in g.

There is 709.1 grams of urea needed.

Explanation:

Step 1: The balanced equation:

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

Step 2: Data given

The exhaust stream of an automobile has a flow rate of 2.37 L/s

This happens at a temperature of 657 Kelvin

The partial pressure of NO is 14.0 torr

Step 3: Calculating the volume during 8.1 hours

V = 2.37L/s ∙ 8.1hr ∙3600s/hr = 69109.2L = 69.1092m³

Step 4: Calculate the partial pressure of nitric oxide:

p = 14 torr ∙ 101325Pa/760torr = 1866.5Pa

Step 5: Calculating number of moles of NO

⇒ We use the ideal gas law P*V=n*R*T

n (NO) = P*V/R*T

with P = The partial pressure of NO = 1866.5 Pa

with V = the volume of NO = 69.1092m³

with R = the gas constant = 8.314472 Pa*m³/mol*K

with T = 657 Kelvin

n (NO) = 1866.5Pa ∙ 69.1092m³ / (8.314472Pa*m³/mol*K ∙ 657K)

= 23.61 mol es

Step 6: Calculate moles of urea

Since there is consumed 2 moles of urea per 4 moles of nitric oxide.

This means for 24.432 moles of NO, there is consumed 23.61 / 2 = 11.806 moles of urea.

Step 7: Calculate mass of urea

m (CO (NH₂) ₂) = 11.806 moles ∙ 60.06g/moles = 709.07g

There is 709.1 grams of urea needed.