A mixture of c2h2 and c3h8 occupied a certain volume at 80mm the mixture was completely burnt into c02 and h2o when the pressure of co2 was found to be 230 mm at same t and v find mole fraction of c3h8
Answers
Mole fraction of C₃H₈ in the mixture is 0.875
Given-
- Pressure occupied initially by C₂H₂ and C₃H₈ = 80 mm of Hg
- Pressure occupied by CO₂ in both the cases = 230 mm of Hg
From the combustion reaction of C₂H₂
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
x 2x
From the combustion reaction of C₃H₈
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
y 3y
From the given data we can write that-
x + y = 80
2x + 3y = 230
By solving the above two equations we get the value of x and y which is
x = 10
y = 70
So mole fraction of C₃H₈ = 70/70+10 = 70/80 = 0.875
Answer:
Explanation:
Pressure is directly proportional to the no. of moles.
Therefore,we can treat pressure as moles.
Now, assume the pressure of C2H2 as X. Therefore, pressure of C3H8 is(80-X)
Now,
1)C3H8 + 5 O2 = 3 CO2 + 4 H20
Initially: (80-X)
Finally: 3(80-X)
2)C2H2 + 5/2 O2 = 2 CO2 + H2O
Initially: X
Finally: 2X
Total pressure of CO2 after reaction=230mmHg
Therefore,
2X + 3(80-X) = 230
After solving, X=10
It means 10 atmC2H2 and 70atm C3H8 were in the mixture initially but,
According to the ideal gas equation (PV=nRT), pressure is directly proportional to the no. of moles when Temperature and Volume are constant
Therefore,The mixture has 10 moles of C2H2 and 70 moles of C3H8
Fraction of C2H2 in mixture=Mole fraction of C2H2 in the mixture
Mole fraction of C2H2:
10/(70+10)
=1/8
=0.125
Therefore, fraction of C2H2 in the mixture= 0.125