The apparatus contains methane and oxygen at a temp of 383K.

Here is the reaction:

CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)

solve for final pressures of all species and the final total pressure,

For CH4 the volume = 1.164L
the pressure = 6atm

For O2 the volume = 0.827L
the pressure = 32atm

So far I have calculated for the moles of ch4 and o2, by the ideal gas law.

I then used the molar ratio to find the moles of product from each individual reactant. I determined that the limiting reactant is CH4 (don't know if that's correct).

From there would I just use the moles of products produced from the excess reactant to determine the pressure of co2 and h20? How do I find the third partial pressure?

You seem to be on the right track. Convert mols of products to pressure for each.

To find the O2 pressure at the end of the reaction, you know moles O2 initially, convert mols CH4 used to mols O2 used, then mols O2 initially - mols O2 used = mols O2 left unreacted. Then change that to pressure.

okay thank you very much!

To solve for the final pressures of all species and the final total pressure, we can use the ideal gas law equation and the stoichiometry of the reaction.

First, let's calculate the final number of moles for each gas using the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L atm / mol K)
T = temperature (in K)

For CH4:
Given:
P(CH4) = 6 atm
V(CH4) = 1.164 L
T = 383 K

n(CH4) = (P(CH4) * V(CH4)) / (R * T)
= (6 atm * 1.164 L) / (0.0821 L atm / mol K * 383 K)
≈ 0.1905 mol

For O2:
Given:
P(O2) = 32 atm
V(O2) = 0.827 L
T = 383 K

n(O2) = (P(O2) * V(O2)) / (R * T)
= (32 atm * 0.827 L) / (0.0821 L atm / mol K * 383 K)
≈ 3.978 mol

Now, considering the balanced equation: CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)

We can see that for every 1 mole of CH4 reacted, we will produce 1 mole of CO2 and 2 moles of H2O. So, the moles of CO2 and H2O produced will be equal to the moles of CH4 reacted.

Therefore, n(CO2) = n(H2O) = n(CH4) = 0.1905 mol

Now, let's calculate the final volume using the ideal gas law equation, assuming all gases behave ideally:

For CO2:
n(CO2) = 0.1905 mol
P(CO2) = ?
V(CO2) = ?
T = 383 K

Rearranging the ideal gas law equation:

V(CO2) = (n(CO2) * R * T) / P(CO2)

Substituting the values:

V(CO2) = (0.1905 mol * 0.0821 L atm / mol K * 383 K) / P(CO2)

Similarly, for H2O:
n(H2O) = 0.1905 mol
P(H2O) = ?
V(H2O) = ?
T = 383 K

V(H2O) = (n(H2O) * R * T) / P(H2O)

Next, we need to calculate the total pressure, which is the sum of the partial pressures of all the gases present.

P(total) = P(CH4) + P(O2) + P(CO2) + P(H2O)

Substituting the given values:

P(total) = 6 atm + 32 atm + P(CO2) + P(H2O)

Now, we have two unknowns: P(CO2) and P(H2O). To solve for them, we need to use the ideal gas law equation again with the calculated volumes for CO2 and H2O:

For CO2:
P(CO2) = (n(CO2) * R * T) / V(CO2)
= (0.1905 mol * 0.0821 L atm / mol K * 383 K) / V(CO2)

For H2O:
P(H2O) = (n(H2O) * R * T) / V(H2O)
= (0.1905 mol * 0.0821 L atm / mol K * 383 K) / V(H2O)

Finally, we substitute the values of V(CO2) and V(H2O) into the equations above and solve for P(CO2) and P(H2O). Once we have these values, we can substitute them back into the equation for P(total) to obtain the final total pressure.