Chemistry, 25.07.2019 01:20 1233042260

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)+2c o2(g)+4h2o(g) suppose that the exhaust stream of an automobile has a flow rate of 2.32 l/s at 670 k and contains a partial pressure of no of 12.1 torr. what total mass (in gram) of urea is necessary to react completely with the no formed during 8.1 hours of driving?

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Answer from: mphan4093

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 = $8.1\times 60\times 60=29160s$

If, in 1 second, the volume of stream flows is 2.32 L

The, in 29160 seconds, the volume of stream flows will be = $\frac{2.32L}{1s}\times 29160s=67651.2L$

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:

PV=nRT

P = pressure = 1 atm

V = volume = 67651.2 L

n = number of moles = ?

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 670 K

Putting values in above equation, we get:

$1atm\times 67651.2=n\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 670K\\\\n=1231.36mol$

Using the equation given by Raoult's law, we get:

$p_A=\chi_A\times P_T$

$p_{NO}$ = partial pressure of NO = 12.1 torr

$\chi_{NO}$ = mole fraction of NO = ?

$P_{T}$ = total pressure of solution = 1 atm = 760 torr (Conversion factor: 1 atm = 760 torr)

Putting values in above equation, we get:

$12.1torr=\chi_{NO}\times 760torr\\\\\chi_{NO}=0.0159$

Mole fraction of nitrogen oxide is written as:

$\chi_{NO}=\frac{\text{Moles of NO}}{\text{Total moles}}$

Mole fraction of NO = 0.0159

Total moles = 1231.6 moles

Putting values in above equation, we get:

$0.0159=\frac{\text{Moles of NO}}{1231.6}\\\\\text{Moles of NO}=19.57mol$

For the given chemical reaction:

$2CO(NH_2)_2(g)+4NO(g)+O_2(g)\rightarrow 4N_2(g)+2CO_2(g)+4H_2O(g)$

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 = $\frac{2}{4}\times 19.57=9.785mol$ of urea.

To calculate the mass of urea, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of urea = 9.785 mol

Molar mass of urea = 60.06 g/mol

Putting values in above equation, we get:

$9.785mol=\frac{\text{Mass of urea}}{60.06g/mol}\\\\\text{Mass of urea}=587.68g$

Hence, the mass of urea that is necessary to react completely is 587.68 g.

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