Question

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.)

Answer 1

Answer:

$m_{Fe_3O_4}=1.7gFe_3O_4$

Explanation:

Hello,

In this case, considering the given reaction:

$3 Fe(s) + 4 H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4 H_2(g)$

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

$Kc=\frac{[H_2]^4}{[H_2O]^4}$

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

$5.1=\frac{(4x)^4}{(0.05M-4x)^4}$

Thus, the change $x$ is obtained as:

$\sqrt[4]{5.1} =\sqrt[4]{[\frac{(4x)}{(0.05M-4x)}]^4}\\\\1.5=\frac{(4x)}{(0.05M-4x)}\\\\x=0.0075M$

Thus, the moles of hydrogen at equilibrium are:

$[H_2]_{eq}=4*0.0075\frac{mol}{L}*1.0L=0.03molH_2$

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

$m_{Fe_3O_4}=0.03molH_2*\frac{1molFe_3O_4}{4molH_2}*\frac{231.533gFe_3O_4}{1molFe_3O_4} \\m_{Fe_3O_4}=1.7gFe_3O_4$

Best regards.