A mixture of ethane (C2H6) and ethene (C2H4) occupies 40 liters at 1.00 atm. and 400 K.

Question

A mixture of ethane $\left( C _{2} H _{6}\right)$ and ethene $\left( C _{2} H _{4}\right)$ ocupies 40 litres at $1.00$ atm and $400 K$. The mixture reacts completely with $130 \cdot 0 g$ 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.

Answer 1

SOLUTION : No. of moles of mixture $=\frac{1 \times 40}{0 \cdot 082 \times 400}=1 \cdot 22$ moles

Let $C_{2} H_{6}=x$ mole, $C_{2} H_{4}=y$ mole; $\therefore \ddot{x}+y=1 \cdot 22$

Reactions :

$C _{2} H _{6}+\frac{7}{2} O _{2} \rightarrow 2 CO _{2}+3 H _{2} O , C _{2} H _{4}+3 O _{2} \rightarrow 2 CO _{2}+2 H _{2} O$

$x$ mole $\frac{7}{2} x$ mole $\quad y$ mole $3 y$ mole

$130 g O_{2}=\frac{130}{32}$ moles $=4.06$ moles

Hence $\frac{7}{2} x+3 y=4 \cdot 06 \quad$ or, $7 x+6 y=8 \cdot 12$

From (I) and (II), $\frac{7 x+6 y}{x+y}=\frac{8 \cdot 12}{1 \cdot 22}$

Subtracting 6 from both sides

$\frac{x}{x+y}=\frac{8 \cdot 12}{1 \cdot 22}-6=\frac{0 \cdot 8}{1 \cdot 22}= 0 \cdot 656$

$\therefore$ Mole fraction of ethene $= 0 \cdot 344$.