Question

An ideal gas mixture of ethane $\left(C_2{H}_6\right)$ and ethene $\left(C_2{H}_4\right)$ occupies 28 litres at 1 atm and 273 K. The mixture reacts completely with 128 g of $O_2$ to produce $C{O}_2$ and $H_{2}O$. The mole fraction of $C_2{H}_6$ in the mixture is:

• A. 0.6
• B. 0.4
• C. 0.5
• D. 0.8

Answer 1

The correct option is B 0.4

An ideal gas mixture of ethane and ethene occupies 28 L at 1 atm pressure and 273 K.
We know that,
1 mole of a gas at STP occupies 22.4 L
So, 28 L will be occupied by= $\frac{28}{22.4}=1.25mol$
Let $n_{ethane}=a$ and $n_{ethene}=b$
so $a+b=1.25$ .......... (i)

Reaction of ethane and ethene with oxygen:
$C_2{H}_6+\frac{7}{2}{O}_2\to 2C{O}_2+3H_{2}O$
1 mole of $C_2{H}_6$ reacts with $\frac{7}{2}$ mole of $O_2$
So, ‘a’ moles of $C_2{H}_6$ (ethane) will react with$=\frac{7}{2}a$ moles of $O_2$

$C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O$
‘b’ moles of $C_2{H}_4$ react with $3b$ moles of $O_2$
Moles of $O_2$ reacting in this mixture$=\frac{7}{2}a+3b=\frac{128gof{O}_2}{molarmassof{O}_2}=4mol$
$\frac{7}{2}\times a+3b=4$ .......... (ii)

On solving equation (i) and (ii) we get
a=0.5 and b=0.75
Mole fraction of ethane $=\frac{0.5}{1.25}=0.4$