# 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 equilibrium Kc is 0.38. There are 50 grams of water...

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 equilibrium Kc is 0.38. There are 50 grams of water left. How do I calculate the mass of the substances at the equilibrium?

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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.