The reaction of iron and water vapor results in an equilibrium reaction, 3 Fe(s) + 4 H2O(g) ¡ê Fe3O4(s) + 4 H2(g) and an equilibrium constant of 4.6 at 850¡ÆC. What is the concentration of Hydrogen present at equilibrium if the reaction is initiated with 28 g of H2O and excess iron, Fe, in a 10.0 L container?

Answer 1

mols H2O = grams/molar mass H2O = approx 1.6 but this is an estimate and you need a more accurate answer. I've estimated all of the others, too, so be sure to redo each step for better accuracy.
M H2O = mols/L = approx 0.16 M.

.....3Fes) + 4H2Og) ==> Fe3O4s) + 4H2g)
I.....xs......0.16M......0.........0
C.....xs......-4x........solid....4x
E.....xs....0.16-4x......solid....4x

K = 4.6 = (H2)^4/(H2O)^4
4.6 = (4x)^4/(0.16-4x)^4
Solve for x = (H2) at equilibrium.
Post your work if you get stuck.

Answer 2

0.184, but I don't know how I got it :( SOS please help