Question

A mixture of ethane $\left(C_2{H}_6\right)$ and ethene $\left(C_2{H}_4\right)$ occupies 40 L 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 $C_2{H}_4$ and $C_2{H}_6$ in the mixture.

• A. $X_{C_2{H}_6}=0.6707,X_{C_2{H}_4}=0.3293$
• B. $X_{C_2{H}_6}=0.3293,X_{C_2{H}_4}=0.6707$
• C. $X_{C_2{H}_6}=0.3707,X_{C_2{H}_4}=0.6293$
• D. Non eof these

Answer 1

The correct option is A $X_{C_2{H}_6}=0.6707,X_{C_2{H}_4}=0.3293$

For $O_2$
$PV=nRT$
or $1\times 40=n\times 0.0821\times 400$
or $n=1.218$

$C_2{H}_6\left(g\right)+\frac{7}{2}{O}_2\left(g\right)⟶2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

$C_2{H}_4\left(g\right)+3O_2\left(g\right)⟶2C{O}_2\left(g\right)+2H_{2}O\left(l\right)$

Let the moles of ethane be $a$.

Volume of O${}_2$ required by ethane$=\frac{7}{2}a$

Volume of ethene$=1.218-a$

Volume of O${}_2$ required by ethene $=3(1.218-a)$

Given

$\frac{7}{2}a+3(1.218-a)=\frac{130}{32}$(moles of oxygen)

or $\frac{7}{2}a+3.654-3a=4.0625$

$(1.218-a)=(1.218-0.817)=0.401$ (moles of ethene)

Mole fraction of ethane$=\frac{0.817}{1.218}=0.6707$

Mole fraction of ethene $=(1-0.6707)=0.3293$