posted by Joseph .
A mixture of ethane and ethene occupied 35.5 L at 1.000 bar and 405 K. This
mixture reacted completely with 110.3 g of O2 to produce CO2 and H2O.
What was the composition of the original mixture? Assume ideal gas
Write balanced equations for the combustion of ethane and ethene. I have
2C2H6 + 7O2 ==> 4CO2 + 6H2O
C2H4 + 3O2 ==> 2CO2 + 2H2O
Use PV = nRT and solve for mols of the ethane + mols ethene.
Let X = mols ethane
and Y = mols ethene
Then X + Y = mols from above. This is equation 1.
For equation 2, convert mols C2H6(that's X) and mols C2H4(that's Y) to mols O2 needed for combustion.
X*(7/2) + Y*(3/1) = 110.3/32
Solve for two equation simultaneously for X and Y and that will give you mols. Since the problem asks for composition, I would think moles is as good an answer as you want; however, you can convert mols of each to L by using PV = nRT and the conditions of the problem. Post your work if you get stuck.