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.
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
behavior
Im pretty sure i use dalton's law of partial pressures but i cant figure out a solution.
3 answers
Another hour gone by working on this and i still can't figure it out. Im totally lost!
got it thanks!