Question

Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition: Calculate the value of the equilibrium constant K_c for this reaction. Round your answer to 2 significant digits.

Answer 1

Complete Question

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Answer:

The equilibrium constant is
`K_c= 2.8*10^(-4)`

Step-by-step explanation:

From the question we are told that

The chemical reaction equation is

`Fe_(2) O_(3)_((s)) + 3H_(2)_((g)) -----> 2Fe_((s)) + 3H_(2) O_((g))`

The voume of the misture is
`V_m = 5.4L`

The molar mass of
`Fe_(2) O_(3)_((s))` is a constant with value of
`M_{Fe_(2) O_(3)_((s)) } = 160g/mol`

The molar mass of
`H_(2)_((g))` is a constant with value of
`H_2 = 2g/mol`

The molar mass of
`H_(2)O` is a constant with value of
`H_2O = 18g/mol`

Generally the number of moles is mathematically given as

`No \ of \ moles \ = (mass)/(molar\ mass)`

For
`Fe_(2) O_(3)_((s))`

`No \ of\ moles = (3.54)/(160)`

`= 0.022125 \ mols`

For
`H_(2)`

`No \ of\ moles = (3.63)/(2)`

`= 1.815 \ mols`

For
`H_(2)O`

`No \ of\ moles = (2.13)/(18)`

`= 0.12 \ mols`

Generally the concentration of a compound is mathematicallyrepresented as

`Concentration = (No \ of \ moles )/(Volume )`

For
`Fe_(2) O_(3)_((s))`

`Concentration[Fe_2 O_3] = (0.222125)/(5.4)`

`= 4.10*10^(-3)M`

For
`H_(2)`

`Concentration[H_2] = (1.815)/(5.4)`

`= 0.336M`

For
`H_(2)O`

`Concentration [H_2O] = (0.12)/(5.4)`

`= 0.022M`

The equilibrium constant is mathematically represented as

`K_c = ([concentration \ of \ product])/([concentration \ of \ reactant ])`

Considering
`H_2O \ for \ product`

And
`H_2 \ for \ reactant`

At equilibrium the

`K_c = (0.022)/(0.336)`

`K_c= 2.8*10^(-4)`