Question

0.804 g sample of iron ore was dissolved in acid. Iron was reduced to +2 state and it required 47.2 mL of 0.112 N $KMnN{O_4}$ solution for titration. Calculate the percentage of iron of $F{e_3}{O_4}$ in the ore.

Answer 1

Hint: The normality is used to measure the concentration of solute in solution. The normality of the solution is defined as the gram equivalent weight of solute dissolved in one liter of solution. The gram equivalent weight is measured by dividing mass by equivalent weight.

Complete step by step answer:
Given data
Mass of iron is 0.804 g.
Volume of potassium nitrate solution is 47.2 mL.
Normality of potassium nitrate is 0.112 N.
During the titration of iron ore dissolved in acid using potassium nitrate, the ferrous is converted to ferric.
The reaction is shown below.
$5F{e^{2 + }} + MnO_4^ - + 8{H^ + } \to 5F{e^{3 + }} + M{n^{2 + }} + 4{H_2}O$
The redox changes in the reaction is shown below.
$Fe \to F{e^{2 + }} + 2{e^ - }$
$F{e^{2 + }} \to F{e^{3 + }} + {e^ - }$
$5e + M{n^{7 + }} \to M{n^{2 + }}$
The normality of the solution is defined as the gram equivalent of solute dissolved in one litre of solution.
The formula of normality is shown below.
$N = \dfrac{{g.eq}}{V}$
Where,
N is the normality
g.eq is the gram equivalent
V is the volume.
The gram equivalent is calculated by dividing weight by the equivalent weight where equivalent weight is calculated by dividing molecular weight by valency.
The milliequivalent $F{e^{2 + }}$ is equal to milliequivalent of $KMn{O_4}$.
m.eq of $F{e^{2 + }}$ = m.eq of $KMn{O_4}$
m.eq is equal to normality multiplied by volume.
Substitute the values in the above equation.
$\Rightarrow \dfrac{w}{{\dfrac{{56}}{2}}} \times 1000 = 47.2 \times 0.112$
$\Rightarrow \dfrac{w}{{\dfrac{{56}}{2}}} \times 1000 = 5.2864$
$F{e^{2 + }} \to F{e^{3 + }} + 1e$
The mass of $F{e^{2 + }}$ is 0.296 g.
The percentage of iron is calculated as shown below.
$\Rightarrow \% Fe = \dfrac{{0.296 \times 100}}{{0.804}}$
$\Rightarrow \% Fe = 36.82\%$
In $F{e_3}{O_4}$, 3 Fe is present.
$3 \times 56g$ Fe = 232 g
The percentage of $F{e_3}{O_4}$is calculated as shown below.
$\Rightarrow \% \;of\;F{e_3}{O_4} = \dfrac{{0.409}}{{0.804}} \times 100$
$\Rightarrow \% \;of\;F{e_3}{O_4} = 50.87\%$
Therefore, the percentage of iron in $F{e_3}{O_4}$in the ore is 50.87%.

Note:
Make sure that during the titration the milliequivalent of the reactant and the titrant should be equal to reach the equivalence point which results in the formation of the product.