A) 0.50, 0.50
B) 0.75, 0.25
C) 0.67, 0.33
D) 0.25, 0.75
E) 0.33, 0.67
Correct Answer: C
Solution :
For a gaseous mixture of\[{{C}_{2}}{{H}_{6}}\]and\[{{C}_{2}}{{H}_{4}}\] \[pV=nRT\] or \[n=\frac{PV}{RT}=\frac{1\times 41}{0.082\times 500}\] or \[n=1\] \[\therefore \]Total mole of\[{{C}_{2}}{{H}_{6}}+{{C}_{2}}{{H}_{4}}=1\]mole Let the mole of\[{{C}_{2}}{{H}_{6}}=x\] then mole of \[{{C}_{2}}{{H}_{4}}=1-x\] \[{{C}_{2}}{{H}_{6}}+\frac{7}{2}{{O}_{2}}\xrightarrow{{}}2C{{O}_{2}}+3{{H}_{2}}O\] \[{{C}_{2}}{{H}_{4}}+3{{O}_{2}}\xrightarrow{{}}2C{{O}_{2}}+2{{H}_{2}}O\] \[\therefore \]Mole of\[{{O}_{2}}\]needed for complete reaction of mixture \[=\frac{7}{2}x+3(1-x)\] \[\therefore \] \[\frac{7}{2}x+3(1-x)=\frac{10}{3}\] Or \[x=\frac{2}{3}\] Thus, mole fraction of\[{{C}_{2}}{{H}_{6}}=\frac{2}{3}=0.67\] and mole fraction of\[{{C}_{2}}{{H}_{4}}=1-\frac{2}{3}=0.33\]
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