Respuesta :
Answer: The mass of urea that is necessary to react completely is 587.68 g.
Explanation:
We are given:
Rate of flow of exhaust stream = 2.32 L/s
Total time of driving = 8.1 hrs
Converting this into seconds:
Total time of driving = [tex]8.1\times 60\times 60=29160s[/tex]
If, in 1 second, the volume of stream flows is 2.32 L
The, in 29160 seconds, the volume of stream flows will be = [tex]\frac{2.32L}{1s}\times 29160s=67651.2L[/tex]
Let us assume that the total pressure is 1 atm.
- So, to calculate the total number of moles, we use the equation given by ideal gas, which is:
[tex]PV=nRT[/tex]
P = pressure = 1 atm
V = volume = 67651.2 L
n = number of moles = ?
R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = 670 K
Putting values in above equation, we get:
[tex]1atm\times 67651.2=n\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 670K\\\\n=1231.36mol[/tex]
- Using the equation given by Raoult's law, we get:
[tex]p_A=\chi_A\times P_T[/tex]
[tex]p_{NO}[/tex] = partial pressure of NO = 12.1 torr
[tex]\chi_{NO}[/tex] = mole fraction of NO = ?
[tex]P_{T}[/tex] = total pressure of solution = 1 atm = 760 torr (Conversion factor: 1 atm = 760 torr)
Putting values in above equation, we get:
[tex]12.1torr=\chi_{NO}\times 760torr\\\\\chi_{NO}=0.0159[/tex]
- Mole fraction of nitrogen oxide is written as:
[tex]\chi_{NO}=\frac{\text{Moles of NO}}{\text{Total moles}}[/tex]
Mole fraction of NO = 0.0159
Total moles = 1231.6 moles
Putting values in above equation, we get:
[tex]0.0159=\frac{\text{Moles of NO}}{1231.6}\\\\\text{Moles of NO}=19.57mol[/tex]
- For the given chemical reaction:
[tex]2CO(NH_2)_2(g)+4NO(g)+O_2(g)\rightarrow 4N_2(g)+2CO_2(g)+4H_2O(g)[/tex]
By Stoichiometry of the reaction:
4 moles of nitrogen oxide reacts with 2 moles of urea.
So, 19.57 moles of nitrogen oxide will react with = [tex]\frac{2}{4}\times 19.57=9.785mol[/tex] of urea.
- To calculate the mass of urea, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of urea = 9.785 mol
Molar mass of urea = 60.06 g/mol
Putting values in above equation, we get:
[tex]9.785mol=\frac{\text{Mass of urea}}{60.06g/mol}\\\\\text{Mass of urea}=587.68g[/tex]
Hence, the mass of urea that is necessary to react completely is 587.68 g.
