Answer:
There is 709.1 grams of urea needed.
Explanation:
<u>Step 1:</u> The balanced equation:
2CO(NH2)2(g)+4NO(g)+O2(g)→4N2(g)+2CO2(g)+4H2O(g)
<u>Step 2:</u> 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
<u>Step 3:</u> Calculating the volume during 8.1 hours
V = 2.37L/s ∙ 8.1hr ∙3600s/hr = 69109.2L = 69.1092m³
<u>Step 4</u>: Calculate the partial pressure of nitric oxide:
p = 14 torr ∙ 101325Pa/760torr = 1866.5Pa
<u>Step 5:</u> 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
<u>Step 6:</u> 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.
<u>Step 7</u>: Calculate mass of urea
m(CO(NH₂)₂) = 11.806 moles ∙ 60.06g/moles = 709.07g
There is 709.1 grams of urea needed.
