posted by Anonymous on .
A mixture of gases is prepared from 76.9 g of O2 and 16.0 g of H2. After the reaction of O2 and H2 is complete, what is the total pressure of the mixture if its temperature is 178 ºC and its volume is 11.0 L? What are the partial pressures of the gases remaining in the mixture?
2H2 + O2 ==> 2H2O
mols H2 = grams/molar mass = about 8 but you need to calculate that as well as all of the other values that follow.
mols O2 = about 4.8. What is the limiting reagent? I think that's H2 and it will produce about 8 mols H2O.
How much O2 will be used. That's
8mols H2 x (1 mol O2/2 mol H2) = 8 x (1/2) = 4 so you will have 0.8 mols O2 remaining. Summary. We have no H2 left, 0.8 mols O2, and we have formed 8 mols H2O. Add total mols and substitute into PV = nRT and solve for Ptotal.
Then calculate mole fraction O2 and H2O.
Then pO2 = XO2 x Ptotal
pH2O = XH2O x Ptotal.