Question

A mixture of ethane and ethene occupies 40 litre at 1.00 atm and at 400 K.The mixture reacts completely with 130 g of $O_2$ to produce $C{O}_2$ and $H_{2}O$. Assuming ideal gas behaviour, calculate the mole fractions of $C_2{H}_6$ and $C_2{H}_4$ in the mixture.

Answer 1

Volume of the mixture at NTP
$=\frac{40\times 1}{400}\times \frac{273}{1}=27.3litre$
Let the volume of ethane = x litre
Volume of ethane = (273-x)litre
Balanced equations:
$\underset{1vol}{C_2{H}_6}+\underset{7/2vol.}{7/2O_2}\to 2C{O}_2+3H_2$
$\underset{1vol.}{C_2{H}_4}+\underset{3vol.}{3O_2}\to 2C{O}_2+2H_{2}O$
Total volume of oxygen required for complete combustion of the mixture is:
$[\frac{7}{2}x+(27.3-x)\times 3]litre$
or $\left[\frac{\frac{7}{2}x+(27.3-x)\times 3}{2}\right]litre$
Mass of oxygen $\left[\frac{\frac{7}{2}x+(27.3-x)\times 3}{2}\right]\times \frac{32}{22.4}$
$130=(x+163.8)\times \frac{16}{22.4}$
$\therefore x=18.2$
Hence, mole fraction of ethane $=\frac{18.2}{27.3}\times 100=66.66$
Mole fraction of ethane $=33.34$