Respuesta :
If the reaction takes place at 1.00 atm and 365°C then 0.0146 gram of NH₃ to have as a requirement per kL of flue gas.
How to find mole fraction ?
To find mole fraction use this expression
P.P = M.F × T.P
where,
P.P = Partial Pressure
M.F = Mole Fraction
T.P = Total pressure
Here,
Given flue gas total pressure (T.P) is 1.00 atm
Given NO Partial Pressure (P.P) = 4.5 × 10⁻⁵
Now,
P.P = M.F × T.P
4.5 × 10⁻⁵ atm = 1.00 atm × M.F
M.F = 4.5 × 10⁻⁵ L
Volume of NH₃ required
[tex]= \frac{4.5 \times 10^{-5}\ L\ NH_{3}}{1 \text{L flue gas}} \times 1000\ \text{L flue gas}[/tex]
= 4.5 × 10⁻² L NH₃
What is Ideal Gas Law ?
The law which states that the pressure of gas is directly commensurable to the volume and temperature of the gas is known as ideal gas law.
It is expressed as
PV = nRT
where,
P = Pressure
V = Volume in liter
n = number of mole of gas
R = Ideal gas constant
T = Temperature in kelvin
T = 365° C
= 365 + 273
= 638 K
Now,
PV = nRT
[tex]1\ \text{atm} \times 4.5 \times 10^{-2} = n \times 0.08206\ \text{L. atm. mol}^{-1}.K^{-1} \times 638 K[/tex]
n = 8.60 × 10⁻⁴ mol
NH₃ to have as a requirement
[tex]= 8.60 \times 10^{-4}\ \text{mol}\ NH_{3} \times \frac{17.03\ g }{1\ \text{mol}}[/tex]
= 0.0146 g
Thus from the above conclusion we can say that 0.0146 gram of NH₃ to have as a requirement per kL of flue gas.
Learn further about Ideal Gas Law here: https://brainly.com/question/25290815
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