A mixture of ethane and ethene occupies 41 L at 1 atm and 500k the mixture reacts completely with 10/3 moles of O2 to produced CO2 and H2O . The mole fraction of ethane and ethene in the mixture are ?
Answers
Hi,
Answer:
The mole fraction of ethane is 0.667 and ethene is 0.334 in the mixture.
Explanation:
Given data:
Pressure, P = 1 atm
Volume, V = 41 L
Absolute Temperature, T = 500 K
To find: mole fraction of ethane and ethene in the mixture
According to the ideal gas law, we have
PV = nRT
Where R is the ideal gas constant = 0.082 L atm/K mol and n is the no of moles of gas
∴ 1 * 41 = n * 0.082 * 500
Or, n = 1
Therefore, the total moles of (ethane + ethene) = 1
Let the no. of moles of ethane ne “n1” and ethene be “n2”.
C2H6 + C2H4 = 1
∴ n1 + n2 = 1 ……(i)
We are given that ethane and ethene reacts completely with 10/3 moles of O2 to produce CO2 and H2O.
C2H6 + 7/2 O2 → 2CO2 + 3H2O
C2H4 + 3 O2 → 2CO2 + 2H2O
From these above chemical equation we can write as
7/2n1 + 3n2 = 10/3 …. (ii)
Subtracting eq. (i) from (ii), we get
n1 = 2/3 = 0.667
Substituting value of n1 in eq. (i), we get
2/3 + n2 = 1
Or, n2 = 1 - 2/3 = 1/3 = 0.334
Mole fraction of ethane = no. of mole of ethane / total moles = 0.667/1 = 0.667
Mole fraction of ethene = no. of mole of ethene / total moles = 0.334/1 = 0.334
Hope this helps!!!!!
Answer:
0.33
Explanation:
Formula for ethane = C2H6C2H6
Formula for ethane = C2H4C2H4
Total mole of (C2H6+C2H4) = 1.2195
PV = nRT
1×41 = n×0.082×500
n1 + n2 = 1 --- 1
C2H6+7/2C2H6 →2CO2+3H2O
C2H4+3O2→2CO2+2H2O
7/2n1+3n2=10/3 --- 2
On solving 1 and 2
n1 =2/3 and n2=1/3
x1=2/3
= 0.67
x2 =1/3
= 0.33
Therefore, the mole fraction of ethane and ethene in the mixture is 0.33.