**Answer:**

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.