A mixture of ethane (C_(2)H_(6)) and ethane (C_(2)H_(4)) occupies 40 litres at 1.00 atm and 400 K. The mixture reacts completely with 130 g of O_(2) to produce CO_(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 mixture of ethane (C_(2)H_(6)) and ethane (C_(2)H_(4)) occupies 40 l

1.	A mixture of ethane (C_(2)H_(6)) and ethane (C_(2)H_(4)) occupies 40 litres at 1.00 atm and 400 K. The mixture reacts completely with 130 g of O_(2) to produce CO_(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.
Answer» Solution :Calculation of total no. of MOLES in the gaseous mixture Applying ideal gas equation, PV = nRT `"1 atm"xx40L= nxx0.0821"L atm K"^(-1)"mol"^(-1)xx400K` `"or"n=(40)/(0.0821xx400)="1.218 MOLE"` Calculation of no. of moles of each component Suppose no. of moles of `C_(2)H_(6)` in the mixture = x Then no. of moles of `C_(2)H_(4)` in the mixture = `1.218-x` ALSO, 130 G of `O_(2)=(130)/(32)" moles = 4.0625 moles"` The reactions for complete combustion of `C_(2)H_(6)` and `C_(2)H_(4)` are `2C_(2)H_(6)+7O_(2) rarr 4CO_(2)+6H_(2)O"...(i)"` ` C_(2)H_(4)+3O_(2) rarr 2CO_(2)+2H_(2)O"...(ii)"` From eqn. (i), no. of moles of `O_(2)` required for complete combustion of x moles of `C_(2)H_(6)=(7)/(2)XX x=3` From eqn. (ii), no. of moles of `O_(2)` required for complete combustion of `(1.218-x)` moles of `C_(2)H_(4)` `=3(1.218-x)` `therefore""3.5x+3(1.218-x)=4.0625` `"or"0.5x=4.0625-3.654=0.4085"or"x=0.8170" mole"` Calculation of mole fractions `"Mole fraction of "C_(2)H_(6)=(.^(n)C_(2)H_(6))/(n_("total"))=(0.817)/(1.218)=0.67" and mole fraction of "C_(2)H_(4)=1-0.67=0.33`