Question

Given,$0.804\;gm$ sample of iron ore containing only $Fe\;$ and $FeO$ was dissolved in acid. Iron oxidises into a $ + 2$ state and it requires $117.20\;ml$ of $0.112\;N\;{K_2}Cr{O_7}$ solution for titration. Calculate the percentage of iron in the ore.

Answer 1

Hint: Iron ores are minerals and rocks from which metallic iron can be extracted, these ores are generally made up of iron oxides. Iron is usually discovered in the form of Magnetite, Hematite, Limonite, Siderite, Goethite. These iron ores vary in colours from dark grey, bright yellow or deep purple to rusty red. We are given a sample of iron ore to find the percentage of iron. It is processed through titration. We will see the reaction and calculate the percentage of iron in the ore.

Complete answer:
Reaction involved in the above process of extraction of iron through titration
$6F{e^{2 + }}\; + \;C{r_2}{{\text{O}}_7}^{2 - }\; + \;14{{\text{H}}^ + }\; \to \;6F{e^{3 + }}\; + \;2C{r^{3 + }}\; + \;7{{\text{H}}_2}{\text{O}}$
Let’s write the given values
Weight of the sample, ${\text{w}}\; = \;0.804\;gm$
Volume of Potassium Dichromate, ${\text{V}}\; = \;117.20\;ml$
Normality of Potassium Dichromate, ${\text{N}}\; = \;0.112$
Now, calculating the molarity of Potassium dichromate which is given by the formula
${\text{M}}\; = \dfrac{{\;{\text{N}}}}{{{\text{n}} - factor}}$
where M is the molarity, N is the Normality of the potassium Dichromate, substituting the values we get
${\text{M}}\; = \dfrac{{\;0.112}}{6}$
${\text{M}}\; = \;0.018\;mol{L^{ - 1}}$
Now calculating the number of moles of Potassium Dichromate from the molarity formula
${\text{M}}\; = \dfrac{{{\text{n}}\;}}{{{\text{V}}\;}}$
${\text{M}} \times {\text{V}}\; = \;{\text{n}}$
 Substituting given value in the above equation we get
${\text{n}} = \dfrac{{\;0.018\; \times \;117.20}}{{1000}}$
${\text{n}} = 0.0021\;mol$
From the reaction we can observe that six moles of Iron react with one mole of Potassium Dichromate so number of moles of Iron will be
${{\text{n}}_{Fe}} = \;6\; \times 0.0021$
${{\text{n}}_{Fe}} = \;0.0131\;mol$
we can now calculate the weight of iron in the sample
${w_{Fe}} = \;0.0131\; \times \;55.85$
${w_{Fe}} = \;0.7331\;{\text{g}}$
Weight of iron is now known to us, calculating Percentage of Iron in Sample
$\dfrac{{0.7331}}{{0.804}}\; \times \;100\; = \;91.1\% $
So, the given sample of iron ore contains $\;91.1\% $ Iron.

Note:
n-factor is given by the valency factor or conversion factor, in redox reactions it is given by the number of moles of electrons gained or lost. There is loss of six moles of electrons in Chromium in this reaction So n-factor of Potassium Dichromate is Six.