A mixture of ethane and ethene occupies 41 L at 1 atm and 500 K. the mixture reacts completely with $\frac{10}{3}$ mole of $O_{2}$ to produce ${CO}_{2}$ and $H_{2}O$. The mole fraction of ethane and ethene in the mixture are $\left( R = 0.082L\ atm\ K^{- 1}\ {mol}^{- 1} \right)$ respectively

Question

Dr. Abubacker · Accepted Answer

(c)
pV = nRT
1$	imes 41 = n 	imes 0.0821 	imes 500\$
$\ \ \ \ \ \ \ \ \ n\  = \ 0.998\$ mol
The no. of moles of ethane = x
so no. of moles of ethane = (0.998 - x)
Reaction of ethane and ethene with $\mathbf{O}_{\mathbf{2}}\mathbf{:}$
(i) ${2C}_{2}H_{6} + 7O_{2} ightarrow 4{CO}_{2} + 6H_{2}O$
(ii) $C_{2}H_{4} + 3O_{2} ightarrow 2{CO}_{2} + 2H_{2}O$
According to (i) reactions
2 mole ethane reacts with =7 mole $O_{2}$
x mole ethane react with = $\frac{7_{x}}{2}$ mole $O_{2}$
According to (ii) reactions
1 mole ethene reacts with = 3 mole $O_{2}$
(0.998 -x)mole ethene reacts =3 (0.998-x)mole of $O_{2}$
$\frac{7_{x}}{2}$ +[3 (0.998-x)]=$\frac{10}{3}$mole of $O_{2}$
3.5x + 2.994 -3x=$\frac{10}{3}$ mole of $O_{2}$
0.5x = 3.333 - 2.994 = 0.3393
x=$\frac{0.3393}{0.5} =$ 0.678 mole of ethane
moles of ethene = 0.998 – 0.678 =0.32

Q. A mixture of ethane and ethene occupies 41 L at 1 atm and 500 K. the mixture reacts completely with $\frac{10}{3}$ mole of $O_{2}$ to produce ${CO}_{2}$ and $H_{2}O$. The mole fraction of ethane and ethene in the mixture are $\left( R = 0.082L\ atm\ K^{- 1}\ {mol}^{- 1} \right)$ respectively

Explanation

Reviewed by Dr. Abubacker

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