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)
Show Answer
Answer:
Correct Answer: 77. $(0.221 atm)$
Solution:
- The total moles of gaseous mixture $=\frac{p V}{R T}=\frac{1 \times 40}{0.082 \times 400}$
$$ =1.22 $$
Let the mixture contain $x$ mole of ethane. Therefore,
$$ \begin{aligned} & \underset{x}{C _2 H _6}+\frac{7}{2} O _2 \longrightarrow 2 CO _2+3 H _2 O \ & C _2 H _4+3 O _2 \longrightarrow 2 CO _2+2 H _2 O \ & 1.22-x \end{aligned} $$
Total moles of $O _2$ required $=\frac{7}{2} x+3(1.22-x)=\frac{x}{2}+3.66$
$$ \Rightarrow \quad \frac{130}{32}=\frac{x}{2}+3.66 $$
$\Rightarrow x=0.805$ mole ethane and 0.415 mole ethene.
$\Rightarrow$ Mole fraction of ethane $=\frac{0.805}{1.22}=0.66$
Mole fraction of ethene $=1-0.66=0.34$
