A mixture of ethane and ethene occupied 35.5 L at 1.000 bar and 405 K. This

Question

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
behavior

Im pretty sure i use dalton's law of partial pressures but i cant figure out a solution.

Answer 1

for the mixture, solve n in PV=nRT. That is the total number of moles of both gases.

Now write two equations
2C2H6+ 702>>4CO2 + 6H2O
C2H4 + 3O2 >> 2CO2 + 2H2O

Now, assume n1 grams of O2 in the first reaction, and 110.3 - n1 grams in the second.

Now assume you used k moles of ethane, and L moles of ethene. You know L+k = total moles n you found first.

Now you know the mole ratios in each reaction. for example, if you reacted k moles of ethane, you must have reacted 7/2 mokes if O2, but that had to equal ni grams of O2.

SO you have now plenty of information, and you can solve it. GEt a large tablet of paper.

Answer 2

Another hour gone by working on this and i still can't figure it out. Im totally lost!

Answer 3

got it thanks!