States of Matter - Result Question 77
77. A mixture of ethane $\left(C _2 H _6\right)$ and ethene $\left(C _2 H _4\right)$ occupies $40 L$ at $1.00 atm$ and at $400 K$. The mixture reacts completely with $130 g$ of $O _2$ to produce $CO _2$ and $H _2 O$. Assuming ideal gas behaviour, calculate the mole fractions of $C _2 H _4$ and $C _2 H _6$ in the mixture.
(1995, 4M)
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Solution:
The total moles of gaseous mixture $=\dfrac{p V}{R T}=\dfrac{1 \times 40}{0.082 \times 400}$
$ =1.22 $
Let the mixture contain $x$ mole of ethane. Therefore,
$ \begin{aligned} & \underset{x}{C _2 H _6}+\dfrac{7}{2} O _2 \longrightarrow 2 CO _2+3 H _2 O \\ & C _2 H _4+3 O _2 \longrightarrow 2 CO _2+2 H _2 O \\ & 1.22-x \end{aligned} $
Total moles of $O _2$ required $=\dfrac{7}{2} x+3(1.22-x)=\dfrac{x}{2}+3.66$
$ \Rightarrow \quad \dfrac{130}{32}=\dfrac{x}{2}+3.66 $
$\Rightarrow x=0.805$ mole ethane and 0.415 mole ethene.
$\Rightarrow$ Mole fraction of ethane $=\dfrac{0.805}{1.22}=0.66$
Mole fraction of ethene $=1-0.66=0.34$
BETA
