Question

A mixture of ethane $\left(C_2{H}_6\right)$ and ethene $\left(C_2{H}_4\right)$ 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 $C_2{H}_6$ in the mixture and write numerical value (upto 2 digit only) after decimal point only (as for 0.19, write 19).

Answer 1

For a gaseous mixture of $C_2{H}_6$ and $C_2{H}_4$
$PV=nRT$
$\therefore 1\times 40=n\times 0.082\times 400,\therefore n=1.2195$
$\therefore$ Total mole of $C_2{H}_6+C_2{H}_4=1.2195$
Let mole of $C_2{H}_6$ and $C_2{H}_4$ be a, b respectively
$a+b=1.2195$ ....(i)
$C_2{H}_6+\left(7/2\right)O_2\to 2C{O}_2+3H_{2}O$
$C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O$
$\therefore$ Mole of $O_2$ needed for complete reaction of mixture
$=\left(7a/2\right)+3b$
$\therefore \frac{7a}{2}+3b=\left(\frac{130}{32}\right)$ ....(ii)
By Eqs. (i) and (ii), $a=0.808,b=0.4115$
Mole fraction of $C_2{H}_6=0.808/1.2195=0.66$ and mole fraction of $C_2{H}_4=0.34$
So, answer is 66.