A mixture of ethane and ethene occupies 41 L at 1 atm and 500 K. The mixture reacts completely with 103 of O2 to produce CO2 and H2O. The mole fraction of ethane and ethane in the mixture ate R=0.082LatmK−1mol−1 respectively

Pv=nRT1×41=n×0.082×500n=1,n1+n2=1C2H6+72O2⟶2CO2+3H2OC2H4+3O2⟶2CO2+2H2O72n1+3n2⟶103n1=23 and n2=13n1=23=0.67,n2=0.33

A mixture of ethane and ethene occupies 41 L at 1 atm and 500 K. The mixture reacts completely with 103 of O2 to produce CO2 and H2O. The mole fraction of ethane and ethane in the mixture ate R=0.082LatmK−1mol−1 respectively

Pv=nRT1×41=n×0.082×500n=1,n1+n2=1C2H6+72O2⟶2CO2+3H2OC2H4+3O2⟶2CO2+2H2O72n1+3n2⟶103n1=23 and n2=13n1=23=0.67,n2=0.33

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

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0.67, 0.33

0.50, 0.50

0.25, 0.75

0.75, 0.25

answer is C.

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Detailed Solution

$\begin{array}{l} Pv = n RT \\ 1 \times 41 = n \times 0.082 \times 500 \\ n = 1, n_{1} + n_{2} = 1 \\ C_{2} H_{6} + \frac{7}{2} O_{2} ⟶ 2 {CO}_{2} + 3 H_{2} O \\ C_{2} H_{4} + 3 O_{2} ⟶ 2 {CO}_{2} + 2 H_{2} O \\ \frac{7}{2} n_{1} + 3 n_{2} ⟶ \frac{10}{3} \end{array}$