Consider the following reaction. 3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g) At 900°C, Kc for the reaction is 5.1. If 0.050 mol of H2O(g) and 0.100 mol of Fe(s) are placed in a 1.0 L container at 900°C, how many grams of Fe3O4 are present when equilibrium is established? (This one is somewhat hard.)

Explanation:

Hello,

In this case, considering the given reaction:

Thus, for the equilibrium, just water and hydrogen participate as iron and iron(II,III) oxide are solid:

Thus, at the beginning, the concentration of water is 0.05 M and consequently, at equilibrium, considering the ICE procedure, we have:

Thus, the change is obtained as:

Thus, the moles of hydrogen at equilibrium are:

Therefore, the grams of iron(II,III) oxide finally result:

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