Q.79..

Q79. An ideal gaseous mixture of ethane (C₂H₆) and ethene (C₂H₄) occupies 28 litre at 1atm and 273 K. The mixture reacts completely with 128 g O₂ to produce CO₂ and H₂O. Mole fraction at C₂H₆ in the mixture is :

   (a) 0.6                           (b) 0.4                     (c) 0.5                         (d) 0.8 
 

Dear Student

As given that the mixture of ethane and ethene occupies 29 litre at 1 atm and 273 K. 
And the mixture reacts completely with 128 g of O₂ to produce CO₂ and H₂O. 
Total amount of mixture of ethane and ethene = $n = \frac{PV}{RT}= \frac{1 \times 28}{0.0831 \times 273}= 1.234 mol$

Amount of oxygen used in complete combustion $= \frac{128}{32}= 4 mol$

The reaction of the ethane with oxygen:

$$\begin{gathered} C_2{H}_6 +\left(\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)O_2\to 2C{O}_2 + 3H_{2}O \\ According to stoichiometry 1 mole requires 7/2 mole of oxygen or 112 g of oxygen \end{gathered}$$

The reaction of the ethene with oxygen:

$$\begin{gathered} C_2{H}_4 +3O_2\to 2C{O}_2 + 2H_{2}O \\ According to stoichiometry 1 mole of ethene requires 3 mole of oxygen or 96 g of oxygen \end{gathered}$$

$$\begin{gathered} So if n_1 and n_2 are the amount of C_2{H}_4 and C_2{H}_6 \\ {n}_1 +\hspace{0.17em}{n}_2 = 1.234 mol ........\left(1\right)\hspace{0.17em} \\ 3 n_1 +\hspace{0.17em}7/2 n_2 = 4 mol .......\left(2\right)\hspace{0.17em} \\ Multiply eq \left(1\right)\hspace{0.17em}by 3 we get \\ 3n_1 +3\hspace{0.17em}{n}_2 = 3.702 mol ..........\left(3\right) \\ \hspace{0.17em} \\ Subtracting eq 3 from 2 we get \\ 0.5 n_2 = 0.298 \\ {n}_2 = 0.596 mol \\ Solving for n_1 we get : \\ {n}_1 = 1.234 - 0.596 \\ = 0.638 mol \end{gathered}$$
So, the mole fraction of C₂H₆ : 

$$\begin{gathered} = \frac{0.638}{0.638 + 0.596} \\ = \frac{0.638}{1.234} \\ = 0.517 \end{gathered}$$

So, the correct answer is (c) 

Regards