Question

A mixture ofn ethane and ethane ocupies 40 litres at 1 atm at 400 k. The mixture reacts completely with 130 gm of $O_2$ to produce $C{O}_2$ and $H_{2}O$. assuming ideal behaviour, calculate the mole fractions of ethane and ethene in the mixture.

• A. Ethane is 80% and ethene is 20%
• B. Ethane is 66% and ethene is 34%
• C. Ethane is 25% and ethene is 75%
• D. Ethane is 34% and ethene is 66%

Answer 1

Answer: (B)

Ethane is 66% and ethene is 34%

Let volume of ethane $=x$ Litres
Volume of ethene $=(40-x)$ Litres as total volume is $40$ Litres.

$C_2{H}_6+C_2{H}_4+\frac{13}{2}{O}_2\to 4C{O}_2+5H_{2}O$
(i) $\underset{\left(x\right)}{C_2{H}_6}+\frac{7}{2}\underset{\left(\frac{7}{2}x\right)}{O_2}\to 2C{O}_2+3H_{2}O$

(ii) $\underset{(40-x)}{C_2{H}_4}+\underset{3(40-x)}{3O_2}\to 2C{O}_2+2H_{2}O$

Total $O_2=\frac{7}{2}x+3(40-x)$ at STP

Now we will calculate total number of moles of $O_2$ at 1.00 atm and 400 K

The ideal gas equation is $PV=nRT$

$n=\frac{PV}{RT}=\frac{1}{RT}[\frac{7}{2}x+3(40-x\left)\right]$

Molar mass of oxygen is 32 g/mol.

Mass of $O_2=\frac{1}{RT}[\frac{7}{2}x+3(40-x\left)\right]\times 32=130$ gm (given)

$\therefore \frac{\left[\frac{7}{2}x+3(40-x)\right]\times 32}{0.082\times 100}=130$

$x=26.5$ Litre

Percentage of $C_2{H}_6=\frac{26.5}{40}\times 100=66.25$%

Percentage of $C_2{H}_4=100-66.25=33.75$%