Answer :
Answer:0.86
Explanation:
A mixture of methane gas, CH4(g), and pentane gas, C5H12(g), has a pressure of 0.5922 atm when placed in a sealed container.Lets suppose initially partial pressure of CH4 is x and C5H12 is y
x+y=0.5922-----i
1.CH4[g] + 2O2[g] -> CO2[g] + 2 H2O[g]
2.C5H12 + 8O2 → 5CO2 + 6H2O
partial pressure of CH4 is x produce xatm of CO2 and 2xatm of H2O
x + 2x = 3 atm
partial pressure of C5H12 is y produce 5yatm of CO2 and 6yatm of H2O
5y+6y=11y
so the final partial pressure of product mixture will be
3x + 11y = 2.429atm ------ii
solve eq i for x and put in ii
x=0.5922-y -----iii
3(0.5922-y)+ 11y = 2.429atm
y= 0.8155atm
put in iii
x=0.5922-0.8155
x= 0.51atm (Partial pressure of CH4)
mol fraction of CH4=Partial pressure/
Total pressure
=0.51/0.5922
=0.86.
The mole fraction of methane, CH₄ in the initial mixture is 0.795
Let the partial pressure of methane, CH₄ be u
Let the partial pressure of pentane, C₅H₁₂ be v
The total pressure of the mixture is 0.5239 atm i.e
Partial pressure of methane + partial pressure of pentane = 0.5239 atm
u + v = 0.5239..... (1)
Combustion reaction of the mixture.
For methane CH₄:
CH₄ + 2O₂ —> CO₂ + 2H₂O
From the balanced equation above,
u atm of CH₄ produced u atm of CO₂ and 2u atm of H₂O.
Thus, the partial pressure of CH₄ produces = u + 2u = 3u atm of the products.
For pentane C₅H₁₂:
C₅H₁₂ + 8O₂ —> 5CO₂ + 6H₂O
From the balanced equation above,
v atm of C₅H₁₂ produced v atm of CO₂ and 6v atm of H₂O.
Thus, the partial pressure of C₅H₁₂ produces = 5v + 6v = 11v atm of the products.
Summary:
CH₄ produced 3u atm
C₅H₁₂ produced 11v atm
But the total pressure of the products is 2.429 atm.
Thus,
3u + 11v = 2.429......... (2)
Next, we shall determine the partial pressure of methane, CH₄ i.e u.
u + v = 0.5239 ......... 1
3u + 11v = 2.429 ........2
From equation 1,
v = 0.5239 – u
Substitute v into equation 2
3u + 11v = 2.429
3u + 11(0.5239 – u) = 2.429
3u + 5.7629 – 11u = 2.429
Collect like terms
3u – 11u = 2.429 – 5.7629
–8u = –3.3339
Divide both side by –8
u = –3.3339 / –8
u = 0.4167 atm
Therefore, the partial pressure of methane, CH₄ in the initial mixture is 0.4167 atm.
Finally, we shall determine the mole fraction of methane CH₄ in the initial mixture.
Partial pressure of methane, CH₄ = 0.4167 atm.
Total pressure = 0.5239 atm
Mole fraction of methane CH₄ =?
Mole fraction = partial pressure / Total pressure
Mole fraction of methane CH₄ = 0.4167 / 0.5239
Mole fraction of methane CH₄ = 0.795
Therefore, the mole fraction of methane CH₄ in the initial mixture is 0.795
Learn more: https://brainly.com/question/15712083