At a fixed temperature, equal moles of H2 (g) and O2 (g) are mixed in a constant

Question

At a fixed temperature, equal moles of H2 (g) and O2 (g) are mixed in a constant pressure container (the volume of the container changes in order to keep the pressure at a constant value). The H2 (g) and O2 (g) are allowed to react, producing H2O (g):

2 H2 (g) + O2 (g) --> 2 H2O (g)

If the initial volume of the container, before any reaction takes place, is 2.50 L, determine the volume of the container after the H2 (g) and O2 (g) have reacted to completion.

What is the final volume?

Hint:

Because P and T are constant, Avogadro's Law applies (V is proportional to n).
For this problem, assume that n moles of H2 (g) react with n moles of O2 (g) (equal moles are reacted). The initial moles of gas present are n + n = 2n.
Next, determine the change in the number of moles of each reactant and product that occur when the reaction goes to completion.
From the initial moles of reactants and product present along with the change in moles that occurs when the reaction goes to completion, determine the final moles of each reactant and product present.
Finally, determine the total moles of gases present after the reaction, then use the relationship: nfinal/ninitial = Vfinal / Vinitial.

Explanation / Answer

Avogadro's equation will be used, which is PV=nRT. There are changing conditions, so (P1V1)/(n1t1)=R=(P2V2)/(n2T2) P is pressure, V is volume, n is number of moles, t is temperature in Kelvin, and R is a universal constant.

The problem states that temperature and pressure are constant. So we can cancel these out as well as R. We are left with (V1)/(n1)=(V2/n2) We now need to find the number of mols.

Since you have equal moles of each, assume that you have 1 mol of both H2 and O2.
The total number of moles then is 2 moles each. Set this as your n1 value.

Find the limiting reagent by taking total number of H2 mols divided by number of mols needed for reaction to take place. (1mol H2 / 2 mol H2 needed) * (1 mol O2 needed) to find the number of O2 mols needed. This would be .5, Since there is 1 mol of O2 present, H2 is limiting.

Take H2 and for every 2 mol H2 there are 2 mols H2O. (1 mol H2/2molH2) *2 mol H2O to get 1 mol H2O. Add the moles of oxygen plus the moles of H2O, to get the total n2 value. (1 mol H2O + .5 O2) = 1.5 = n2.

Now we can solve:
(V1/n1)=(V2/n2)

(V1n2)/(n1) = V2

(4.10 L)(1.5 mol) / 2 = 3.075 L

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