Respuesta :
Answer: The concentration of [tex]Fe^{2+}[/tex] ions is [tex]0.548gFe^{2+}/g\text{ steel}[/tex]
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of [tex]KMnO_4[/tex] = 0.140 M
Volume of solution = 34.59 mL
Putting values in above equation, we get:
[tex]0.140M=\frac{\text{Moles of }KMnO_4\times 1000}{34.59}\\\\\text{Moles of }KMnO_4=\frac{0.140\times 34.59}{1000}=4.84\times 10^{-3}mol[/tex]
The chemical equation for the reaction of iron (II) ion with potassium permanganate follows:
[tex]5Fe^{2+}+MnO_4^-+8H^+\rightarrow 5Fe^{3+}+Mn^{+2}+4H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of permanganate ions react with 5 moles of iron (II) ions.
So, [tex]4.84\times 10^{-3}mol[/tex] moles of permanganate ions will react with = [tex]\frac{5}{1}\times 4.84\times 10^{-3}=2.42\times 10{-2}mol[/tex] of iron (II) ions.
To calculate the mass for given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of iron (II) ions = 55.845 g/mol
Moles of iron (II) ions = [tex]2.42\times 10^{-2}[/tex] moles
Putting values in above equation, we get:
[tex]2.42\times 10^{-2}mol=\frac{\text{Mass of }Fe^{2+}\text{ ions}}{55.845g/mol}\\\\\text{Mass of }Fe^{2+}\text{ ions}=(2.42\times 10^{-2}mol\times 55.845g/mol)=1.351g[/tex]
To calculate the concentration of [tex]Fe^{2+}[/tex] ions in steel by mass, we use the equation:
[tex]\text{Concentration of }Fe^{2+}\text{ ions}=\frac{\text{Mass of }Fe^{2+}\text{ ions}}{\text{Mass of steel}}[/tex]
Mass of steel = 2.465 g
Mass of [tex]fe^{2+}[/tex] ions = 1.351 g
Putting values in above equation, we get:
[tex]\text{Concentration of }Fe^{2+}\text{ ions}=\frac{1.351g}{2.465g}=0.548[/tex]
Hence, the concentration of [tex]Fe^{2+}[/tex] ions is [tex]0.548gFe^{2+}/g\text{ steel}[/tex]
