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A mixture of ethane $(C_2H_6)$ and ethene $(C_2H_4)$ occupies 40 litre at 1.00atm and at 400k.The mixture reacts completely with 130g of $O_2$ to produce $CO_2$ and $H_2O$.Assuming ideal gas behavior ,Calculate the mole fractions of $C_2H_4$ and $C_2H_6$ in the mixture.

$\begin{array}{1 1}(a)\;1.2,3.2&(b)\;0.66,0.34\\(c)\;0.1,0.2&(d)\;2.2,3.1\end{array}$

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Mixture of $C_2H_6$ and $C_2H_4$
$PV=nRT$
$1\times 40=n\times 0.082\times 400$
Total mole of $(C_2H_6+C_2H_4)=1.2195$
$C_2H_6+C_2+H_4=1.2195$
$a+b=1.2195$-------(1)
$C_2H_6+\large\frac{7}{2}$$O_2\rightarrow 2CO_2+3H_2O$
$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$
$\large\frac{7a}{2}$$+3b=\large\frac{130}{32}$------(2)
From (1) & (2)
$a=0.808$
$b=0.4115$
Mole fraction of $C_2H_6=\large\frac{0.808}{1.2195}=$$0.66$
Mole fraction of $C_2H_4=0.34$
Hence (b) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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