Question

The Fe2 (55.845 g/mol) content of a 2.465 g steel sample dissolved in 50.00 mL was determined by tiration with a standardized 0.140 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 34.59 mL to reach the end point. What is the concentration of iron in the steel sample? Express your answer as grams of Fe per grams of steel (g Fe2 / g steel).

Answer 1

Answer: The concentration of
`Fe^(2+)` ions is
`0.548gFe^(2+)/g\text{ steel}`

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}`

Molarity of
`KMnO_4` = 0.140 M

Volume of solution = 34.59 mL

Putting values in above equation, we get:

`0.140M=\frac{\text{Moles of }KMnO_4* 1000}{34.59}\\\\\text{Moles of }KMnO_4=(0.140* 34.59)/(1000)=4.84* 10^(-3)mol`

The chemical equation for the reaction of iron (II) ion with potassium permanganate follows:

`5Fe^(2+)+MnO_4^-+8H^+\rightarrow 5Fe^(3+)+Mn^(+2)+4H_2O`

By Stoichiometry of the reaction:

1 mole of permanganate ions react with 5 moles of iron (II) ions.

So,
`4.84* 10^(-3)mol` moles of permanganate ions will react with =
`(5)/(1)* 4.84* 10^(-3)=2.42* 10{-2}mol` of iron (II) ions.

To calculate the mass for given number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Molar mass of iron (II) ions = 55.845 g/mol

Moles of iron (II) ions =
`2.42* 10^(-2)` moles

Putting values in above equation, we get:

`2.42* 10^(-2)mol=\frac{\text{Mass of }Fe^(2+)\text{ ions}}{55.845g/mol}\\\\\text{Mass of }Fe^(2+)\text{ ions}=(2.42* 10^(-2)mol* 55.845g/mol)=1.351g`

To calculate the concentration of
`Fe^(2+)` ions in steel by mass, we use the equation:

`\text{Concentration of }Fe^(2+)\text{ ions}=\frac{\text{Mass of }Fe^(2+)\text{ ions}}{\text{Mass of steel}}`

Mass of steel = 2.465 g

Mass of
`fe^(2+)` ions = 1.351 g

Putting values in above equation, we get:

`\text{Concentration of }Fe^(2+)\text{ ions}=(1.351g)/(2.465g)=0.548`

Hence, the concentration of
`Fe^(2+)` ions is
`0.548gFe^(2+)/g\text{ steel}`