States of Matter 1 Question 77
77. A mixture of ethane $\left(\mathrm{C}{2} \mathrm{H}{6}\right)$ and ethene $\left(\mathrm{C}{2} \mathrm{H}{4}\right)$ occupies $40 \mathrm{~L}$ at $1.00 \mathrm{~atm}$ and at $400 \mathrm{~K}$. The mixture reacts completely with $130 \mathrm{~g}$ of $\mathrm{O}{2}$ to produce $\mathrm{CO}{2}$ and $\mathrm{H}{2} \mathrm{O}$. Assuming ideal gas behaviour, calculate the mole fractions of $\mathrm{C}{2} \mathrm{H}{4}$ and $\mathrm{C}{2} \mathrm{H}_{6}$ in the mixture.
(1995, 4M)
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Answer:
Correct Answer: 77. $(0.221 \mathrm{~atm})$
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,
$$ \begin{aligned} & \underset{x}{\mathrm{C}{2} \mathrm{H}{6}}+\frac{7}{2} \mathrm{O}{2} \longrightarrow 2 \mathrm{CO}{2}+3 \mathrm{H}{2} \mathrm{O} \ & \mathrm{C}{2} \mathrm{H}{4}+3 \mathrm{O}{2} \longrightarrow 2 \mathrm{CO}{2}+2 \mathrm{H}{2} \mathrm{O} \ & 1.22-x \end{aligned} $$
Total moles of $\mathrm{O}_{2}$ 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 $=1-0.66=0.34$
