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.

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.

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

got it thanks!

To solve this problem, you can indeed use Dalton's Law of Partial Pressures. The given information includes the volume of the mixture, the pressure, and temperature. However, the problem does not provide any information about the partial pressures of the individual gases in the mixture, so we need to find a way to determine it.

To apply Dalton's Law of Partial Pressures, we need to know the mole fraction or the number of moles of each component in the mixture. Since the problem does not directly provide this information, we need to use the given reaction and the given amount of O2 to determine the composition of the original mixture.

First, let's balance the chemical equation for the reaction:
C2H6 + C2H4 + O2 → CO2 + H2O

The balanced equation shows that one mole of ethane (C2H6) reacts with one mole of ethene (C2H4) and one mole of oxygen gas (O2) to produce one mole of carbon dioxide (CO2) and one mole of water (H2O).

Next, we need to calculate the amount of O2 using its mass. The molar mass of O2 is 32 g/mol, so we can calculate the number of moles of O2:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 110.3 g / 32 g/mol

Now, we can use stoichiometry to determine the number of moles of ethane and ethene that reacted. Since the reaction is balanced in a 1:1:1 ratio, the number of moles of O2 consumed is equal to the number of moles of ethane and ethene that reacted.
moles of O2 = moles of ethane + moles of ethene

Now that we know the number of moles of O2, we can use the ideal gas law to calculate the volume occupied by O2. The ideal gas law equation is given as:
PV = nRT

We can rearrange the equation to solve for the volume:
V = nRT / P

where:
V = volume
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature
P = pressure

Using the given values (T = 405 K, P = 1.000 bar), we can calculate the volume of O2:
V = moles of O2 * R * T / P

Now that we have the volume of O2, we can subtract it from the total volume of the mixture to determine the volume of the original mixture:
volume of original mixture = total volume - volume of O2

Finally, we can divide the volume of each component by the volume of the original mixture to determine their respective mole fractions. The mole fraction can be calculated using the following equation:
mole fraction = moles of component / total moles

By calculating the mole fraction for ethane and ethene, you can determine the composition of the original mixture.