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Consider the following reaction:

3Fe(s)+4H2O(g)Fe3O4(s)+4H2(g)

(a) What is the apparent oxidation state of Fe in Fe3O4?

(b) Actually, Fe has two oxidation states in Fe3O4. What are they?

(c) At role="math" localid="1654929041124" 9000C, Kc for the reaction is 5.1. If 0.050 mol ofH2O(g) and 0.100 mol of Fe(s) are placed in a 1.0-L container at 9000C, how many grams of role="math" localid="1654929174865" Fe3O4are present at equilibrium?

Short Answer

Expert verified

(a) The apparent oxidation state of Fe inFe3O4 is 2.67.

(b) Fe3O4Can be written asFeO.Fe2O3

Hence oxidation state of F in FeO = +2 and Fe in =+3

(c) 2.42 grams of Fe3O4are present at equilibrium.

Step by step solution

01

Step 1: (a) Apparent oxidation state of Fe in Fe3O4 and (b) two oxidation states of Fe in Fe3O4.

(a) The apparent oxidation state of Fe in Fe3O4is:

Fe3O4=3Fe+4-2=0Fe=83=2.67

(b)localid="1654931375767" Fe3O4Can be written as FeO.Fe2O3

Hence oxidation state of F in FeO = +2 and Fe in =+3

02

Step 2: (c) How many grams of Fe3O4 are present at equilibrium?

Concentrationofwater=MolesVolume[H2O]=0.050mol1.0L=0.050[Fe]=0.100mol1.0L=0.100M

Putting these values in an ICE table:


KC=[Fe3O4][H2]4[Fe]3[H2O]4

Substituting, we get

5.1=(x)(4x)4(0.10-3x)3(0.050-4x)4)5.1=(2556x5)(0.10-3x)3(0.050-4x)4

Since x<< 1

Hence;0.10-3x0.10,&0.05-4x0.050

Hence;

5.1=256x5(0.10-3x)3(0.050-4x)4=256x56.2510-9256x5=3.1910-10

Thus,

x5=1.24510-10x=1.04510-2M

So, we have 1.04510-2mol of Fe3O4 present in 1L of solution.

MassofFe3O4presentatequilibrium=molesmolarmass=1.04510-12232gmol-1=2.42g

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