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JEE Advanced 2021 · Paper 2, Chemistry question 11

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Question Stem for Question Nos. 11 and 12

A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnOX4\ce{KMnO4} solution to reach the end point. Number of moles of FeX2+\ce{Fe^{2+}} present in 250 mL solution is x×10−2\mathbf x \times 10^{-2} (consider complete dissolution of FeClX2\ce{FeCl2}). The amount of iron present in the sample is y\mathbf y% by weight.

(Assume: KMnOX4\ce{KMnO4} reacts only with FeX2+\ce{Fe^{2+}} in the solution

Use: Molar mass of iron as 56 g mol−1^{-1})

The value of x\mathbf x is ____.

Answer and worked solution

Official answer: 1.87

Moles of permanganate. 0.0125×0.03=3.75×10−40.0125\times0.03=3.75\times10^{-4} mol.

Moles of FeX2+\ce{Fe^{2+}}. MnOX4X−+5 FeX2++8 HX+→MnX2++5 FeX3++4 HX2O\ce{MnO4^- + 5Fe^{2+} + 8H+ -> Mn^{2+} + 5Fe^{3+} + 4H2O}, so the 25 mL aliquot holds 5×3.75×10−4=1.875×10−35\times3.75\times10^{-4}=1.875\times10^{-3} mol.

In 250 mL. 1.875×10−21.875\times10^{-2} mol, so x=1.875x=1.875.

Topics:Redox titrationsStoichiometry
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