Question

An ideal gaseous mixture of ethane and ethene occupy $28L$ as STP. The mixture required $128g$ $O_2$ combustion, ,mole fraction of ethene in the mixture is:

• A. $0.4$
• B. $0.5$
• C. $0.6$
• D. $0.8$

Answer 1

The correct option is A $0.4$

In the mixture of ethane and ethene i.e $C_2{H}_6$ and $C_2{H}_4$

As per the ideal gas equation,

$PV=nRT$

$1atm\times 28L=n\times 0.082\times 273K$

No. of moles($C_2{H}_6$ and $C_2{H}_4$)=1.25

let no. of moles of $C_2{H}_6$ and $C_2{H}_4$ be a and b

therefore, a+b=1.25 ....(1)

$C_2{H}_6+\frac{7}{2}{O}_2\to 2C{O}_2+3H_{2}O$

$C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O$

$\frac{7}{2}a+3b=\frac{128}{32}=4$ .....(2)

therefore, solving the equations simultaneously we get

$a=0.5$ and $b=0.75$

Therefore, mole fraction of $C_2{H}_6\left(a\right)$ = $\frac{0.5}{1.25}=0.4$