Answer :
Answer:
[tex]m_{Fe_3O_4}=1.7gFe_3O_4[/tex]
Explanation:
Hello,
In this case, considering the given reaction:
[tex]3 Fe(s) + 4 H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4 H_2(g)[/tex]
Thus, for the equilibrium, just water and hydrogen participate as iron and iron(II,III) oxide are solid:
[tex]Kc=\frac{[H_2]^4}{[H_2O]^4}[/tex]
Thus, at the beginning, the concentration of water is 0.05 M and consequently, at equilibrium, considering the ICE procedure, we have:
[tex]5.1=\frac{(4x)^4}{(0.05M-4x)^4}[/tex]
Thus, the change [tex]x[/tex] is obtained as:
[tex]\sqrt[4]{5.1} =\sqrt[4]{[\frac{(4x)}{(0.05M-4x)}]^4}\\\\1.5=\frac{(4x)}{(0.05M-4x)}\\\\x=0.0075M[/tex]
Thus, the moles of hydrogen at equilibrium are:
[tex][H_2]_{eq}=4*0.0075\frac{mol}{L}*1.0L=0.03molH_2[/tex]
Therefore, the grams of iron(II,III) oxide finally result:
[tex]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[/tex]
Best regards.
The amount of Fe₃O₄ present will be "1.736 g".
Equilibrium
According to the question,
The reaction is:
→ 3Fe (s) + 4H₂O (g) [tex]\rightleftharpoons[/tex] Fe₃O₄ (s) + 4H₂ (g)
Mol of Fe (s) = 0.100
Mol of H₂O (h) = 0.050
We know that the equilibrium be:
→ Kc = [tex]\frac{[H_2]^4}{[H_2O]^4}[/tex]
By substituting the values, we get
5.1 = [tex]\frac{x^4}{(0.050-x)^4}[/tex]
√5.1 = [tex]\frac{x}{(0.050-x)}[/tex]
1.503 = [tex]\frac{x}{(0.050-x)}[/tex]
By applying cross-multiplication, we get
x = 0.030
Now,
The no. of moles Fe₃O₄ produced,
= [tex]\frac{0.030}{4}[/tex]
= 0.0075 mole
hence,
The Fe₃O₄ present be:
= 231.5326 × 0.0075
= 1.736 g
Thus the above answer is correct.
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