A 5.00L reaction vessel contains hydrogen at a partial pressure of 0.588atm and oxygen gas at a partial pressure of 0.302atm. The equation is 2H2(g)+O2(g)-> 2H2O(g). Suppose the gas mixture is ignited and the reaction produces the theoretical yield of the product. What would be the partial pressure of each substance present in the final mixture?

Answer 1

figure the moles of each reaction gas from
PV=nRT solve for n in each case.

Then you have a limiting reactants problem moles H2 and Moles O2. In a complete reaction, you should have 2 moles of H2 to each mole of O2. Examine your ration. If it is higher than that, then oxygen is limiting, so you figure the mole of steam based on the moles of oxygen. For each mole of oxygen you consumed, you consumed twice that amount of hydrogen, and made twice that amount of steam.
Now figure the moles of hydrogen left, the amount you started with, minus the amount consumed. So you have moles of hydrogen, oxygen (zero), steam

from these, calculate the pressure of each gas P=nRT/V

NOW, if the ratio of hydrogen to oxygen is less than two, your steam will be based on the moles of hydrogen, and you will have excess oxygen left, Proceed to calcualte the partial pressures as laid out above