I have a 6 liter container and I put iron and water in it: 3Fe + 4H2O > Fe3O4 + 4H2 This is an equilibrium reaction and the constant of...

For the given well balanced reaction:

`3Fe + 4H_2O -> Fe_3O_4 + 4H_2`

the equilibrium constant can be written as:

`K_c = 0.38 = ([Fe_3O_4][H_2]^4)/([Fe]^3 [H_2O]^4)`

where, [] sign indicates the molar concentration of each substance in mol/l.

Let us assume that we started with "x" g of water. Since 50 g of water is left after the equilibrium has been established, the amount of water that has reacted is x - 50 g.

Since...

For the given well balanced reaction:

`3Fe + 4H_2O -> Fe_3O_4 + 4H_2`

the equilibrium constant can be written as:

`K_c = 0.38 = ([Fe_3O_4][H_2]^4)/([Fe]^3 [H_2O]^4)`

where, [] sign indicates the molar concentration of each substance in mol/l.

Let us assume that we started with "x" g of water. Since 50 g of water is left after the equilibrium has been established, the amount of water that has reacted is x - 50 g.

Since the molar mass of water is 18 g/mole (= 2 x 1 + 16), the moles of water that have been consumed are (x-50)/18. Since, the container has a volume of 6 l, the molar concentration of water that has reacted is given as (x-50)/(18x6) moles/l.

Using stoichiometry, 4 moles of water generate 4 moles of H2. Thus the concentration of H2 is (x-50)/(18x6) mol/l. 

4 moles of water generates 1 mole of Fe3O4. Thus the concentration of Fe3O4 is (x-50)/(18x6x4) mol/l.

And the moles of Fe consumed are (3/4) x [(x-50)/(18x6)] mol/l.

Substituting all these values into the equation for equilibrium constant and solving,  

we get, x = 184.87 g

The moles of water consumed = (x-50)/18 = 7.493 moles of water

Moles of Fe consumed are 3/4 x 7.493 = 5.62 moles. Using the atomic mass of iron (55.85 g/mole), amount of Fe consumed is 313.85 g. 

Thus, the moles of Fe3O4 generated are (x-50)/(18x6x6) = 1.873 moles. Since the molar mass of Fe3O4 is 231.55 g (= 3 x 55.85 + 4 x 16), the amount of Fe3O4 produced is 433.69 g.

Similarly, moles of hydrogen generated are the same of moles of water consumed and thus, are equal to 7.493 moles. Since the molar mass of H2 is 2 g/mole, the amount of H2 generated is 14.986 g or about 15 g.

Hope this helps.