Question

An ideal gaseous mixture of ethane and ethene occupies 23 litre at 1 atmosphere and 273 kelvin the mixture reacts completely with 128 gram o2 to produce co2 and h2o mole fraction at c2 h4 in the mixture is

Answer 1

Answer : The mole fraction of ethene in gaseous mixture are, 0.592

Explanation :

First we have to calculate the moles of gaseous mixture of ethane and ethene.

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 23 L

T = temperature of gas = 273 K

n = number of moles of gaseous mixture = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (23L)=n\times (0.0821L.atm/mol.K)\times (273K)$

$n=1.026mole$

Thant means,

$n_{C_2H_6}+n_{C_2H_4}=1.026$

Let the moles of $C_2H_6$ and $C_2H_4$ be, 'a' and 'b' respectively.

$a+b=1.026$ ............(1)

The balanced chemical reactions are:

(a) $C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

(b) $C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$

From both the reaction, we conclude that

$\frac{7a}{2}+3b=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}$

$\frac{7a}{2}+3b=\frac{128}{32}$ ...........(2)

Now by solving the equation 1 and 2, we get:

a = 0.419

b = 0.607

Now we have to calculate the moles fraction of ethane and ethene in gaseous mixture.

$\text{Mole fraction of }C_2H_6=\frac{\text{Moles of }C_2H_6}{\text{Moles of }C_2H_6+\text{Moles of }C_2H_4}=\frac{0.419}{0.419+0.607}=0.408$

$\text{Mole fraction of }C_2H_4=\frac{\text{Moles of }C_2H_4}{\text{Moles of }C_2H_6+\text{Moles of }C_2H_4}=\frac{0.607}{0.419+0.607}=0.592$

Therefore, the mole fraction of ethene in gaseous mixture are, 0.592