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ValentinkaMS [17]
3 years ago
6

At a certain temperature K = 1.1 X 10^3 L/mol for the reaction given below. Calculate the concentrations of Fe3+, SCN?, and FeSC

N2+ at equilibrium if 0.020 mol of Fe(NO3)3 is added to 1.0 L of 0.10 M KSCN. (Neglect any volume change.)Fe3+(aq) + SCN ?(aq) FeSCN2+(aq)Fe3+____________MSCN- ______________MFeSCN2+ ____________M
Chemistry
1 answer:
lianna [129]3 years ago
6 0

Answer:

Explanation:

Fe⁺³ + SCN⁻ =  FeSCN²⁺

x           x                x

If x mole of Fe⁺³ reacted , unreacted Fe⁺³ = .02 - x

Remaining SCN⁻  unreacted = .1 - x

K_c =   [ FeSCN²⁺] / [Fe⁺³] [ SCN⁻]

1.1 x 10³ =  x  / ( .02 - x ) (.1 - x )

1100 = x / .002 - .12x ( neglecting x² )

x = .0165

Concentration of Fe⁺³ = .02 - x = .0035 M

Concentration of SCN⁻ = .1 - x = .0835 M

Concentration of FeSCN²⁺ = x = .0165M

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The question is incomplete, here is the complete 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 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

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<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

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