Question

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

Answer 1

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?

Answer 2

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.

Answer 3

okay thank you very much!