Question

An ideal gaseous mixture of ethane and ethene occupy 28L as STP. The mixture required 128g ${O_2}$ for combustion, mole fraction of ethane in the mixture is?
(A) 0.4
(B) 0.5
(C) 0.6
(D) 0.8

Answer 1

First try to calculate the moles of gaseous mixture of ethane and ethene, after which try calculating the mole fraction of ethene in gaseous mixture. Formula to find the mole fraction of a compound is as below.
${\text{Mole fraction = }}\dfrac{{{\text{Number of moles of the given compound}}}}{{{\text{Total number Moles}}}}$

Complete step by step answer:
In order to answer this question, first let us try to understand the concept of ideal gas law.
Ideal gas equation: Generally the relation between the temperature, the pressure, and the amount of a gas can be combined into an ideal gas law.

PV = nRT

Where,
P = pressure
V = volume
n = amount of substance
R = ideal gas constant
T = temperature

Keep in mind that standard temperature and pressure (STP) is 0 and 1atm.
The volume of an ideal gas at STP is 22.41L, which is the standard molar volume.

From the given question,
P = pressure of gas = 1atm
V = volume of gas = 28L
T = temperature of gas = ${0^ \circ }C$ = 273+0 = 273K

Now, we need to find the value of n.

N = number of moles of gaseous mixture =?

Here R is the gas constant which is 0.082$Jmo{l^{ - 1}}{K^{ - 1}}$

Now, let us put all the given values in the ideal gas equation,

Therefore, we get
$(1)(28) = n(0.082)(273)$
So, we can write that
$n = \dfrac{{28}}{{(0.082)(273)}} = 0.25$

Therefore, the number of moles of the gaseous mixture is 1.25
Let the number of moles of Ethane and Ethene be a and b.
Hence, a+b = 1.25
Therefore, the balanced chemical reactions are:
${C_2}{H_6} + 3.5{O_2} \to 2C{O_2} + 3{H_2}O$
${C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O$
From both the reactions we can conclude that,
$\dfrac{7}{2}a + 3b = \dfrac{{128}}{{32}}$
Hence, $\dfrac{7}{2}a + 3b = 4$
Therefore, solving the equations simultaneously for a and b, we get,
a = 0.5 and b = 0.75
Hence, the mole fraction of Ethane
${n_{ethane}} = \dfrac{{{\text{Number of moles of ethane}}}}{{{\text{Number of total moles}}}} = \dfrac{{0.5}}{{1.25}} = .4$
Therefore, mole fraction of ethane in the mixture is 0.4

Hence, option A is the required answer.

Note: You should know that the significant deviations from ideal gas behavior generally tend to occur at low temperatures and at very high pressures. The ideal gas can also be used to calculate the density of a gas if the molar mass is known.