[demonat0r]

A mixture of H2(g), O2(g), and 2 millilitres of H2O(l) is present in a 0.500 litre rigid container at 25 degree C. The number of moles of H2 and the number of moles of O2 are equal. The total pressure is 1,146 millimetres mercury. (The equilibrium vapor pressure of pure water at 25 degree C is 24 millimetres mercury.)
The mixture is sparked, and H2 and O2 react until one reactant is completely consumed.
(a) Identify the reactant remaining and calculate the number of moles of the reactant remaining.

Answer:
(a)   2 H2 + O2 ~~> 2 H2O
   mol H2 = mol O2 initially, but 2 mole H2 react for every 1 mol of O2, therefore, O2 is left.
   PT = PH2 + PO2 + PH2O
   1146 mm Hg = PH2 + PO2 + 24 mm Hg
   PH2 + PO2 = 1122 mm Hg
   1122 mm Hg / 4 = PO2 left (1/2 of initial PO2 which is 1/2 total)
   PO2 = 280.5 mm Hg

i do not understand why u have to divide 1122mm Hg by 4. where did they get that number from?
 

[macman104]

You want to find the P_{O2, remain}

In order to do that, you set up an equation describing the initial conditions.

That is your P_tot = P_O2 + P_H2 + P_H2O
Now, you know what P_tot is, so you plug that in , and you know what the P_H2O is as well.

Now, you get an expression that is what the sum of P_O2 and P_H2 is.  That is:

P_O2 + P_H2 = 1122 mm Hg

This is where they make a jump, so I'll explain it:

We solve for P_O2, which we will call x (partly because I'm tired of writing it , and it'll make symbolic manipulation easier), and we'll call P_H2 y, to stay consistent.

We first recognize that x = y, remember the moles of O₂ were equal to moles of H₂.  Combine:
x + x = 1122 mm Hg, to:
2x = 1122 mm Hg, so we see that x or the P_{O2, initial} is equal to 561 mm Hg.

Now we get to the stoichiometry, the balanced equation for the combustion is:

2H₂ + O₂ 2 H₂O.

So, if moles of H₂ = moles  of O₂, let's use 1 because then, that one mole will equal the pressure, and it'll make math easy!

The limiting reagent is H₂, and you'll work out that you will have 0.5 moles of O₂ remaining.  So you get that:

P_{O2, remain} = P_{O2, initial}/2

We divided x (P_O2) by 2 twice, so /4, which is what they did. (Once when we divded x by 2 to get the P_{O2, initial}, and once when we solved had 0.5 moles remaining for P_{O2, remain})

Sorry for the long-windedness, but I hope it was clear!




P.S.  And let me tell you how fun it was to do those subscripts!

[Vidya]

       2H2 + O2 -----  2H2O
          2moles of H2 reacts with 1 mole of O2
         let us say it has x moles of each
     x moles of H2 reacts with x/2 moles of O2
x/2 moles of O2 are left so it makes Hydrogen gas as limiting reactant.
   2 H2 + O2 ~~> 2 H2O
 Before combustion
   PT = PH2 + PO2 + PH2O
   1146 mm Hg = PH2 + PO2 + 24 mm Hg
   PH2 + PO2 = 1122 mm Hg
now initialy no of moles of O2 and H2 (before combustion ) is same
so their partial pressures are also same
so pO2=pH2
so PH2 + PO2 = 1122 mm Hg
it becomes 2pO2=1122mm Hg
    so pO2 (before combustion with x moles)=1122mm Hg /2
after reaction only half moles are left(x/2)
so after reaction pressure due to left O2 gas = pO2/2= 1122mm Hg /4
now we have volume ,pressure and temperature so use ideal gas equation
PV=nRT
n=PV/RT
n will give you no of moles of O2