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Standard XII

Chemistry

Balancing Redox Reaction

A sample 5.6 ...

Question

A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 $m L$ of solution. Titration of 25.0 $m L$ of this solution requires $12.5 m L of 0.03 M K M n O_{4}$ solution to reach the end point. Number of moles of $F e^{2 +}$ present in 250 $m L$ solution is $x \times 10^{- 2}$ (consider complete dissolution of $F e C l_{2} .$ ) The amount of iron present in the sample is $y %$ by weight.
(Assume : $K M n O_{4}$ reacts only with $F e^{2 +}$ in the solution
Use : Molar mass of iron as $56 g m o l^{- 1}$ )

The value of $x$ is .

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Solution

$meq of F e^{+ 2} = meq of K M n O_{4}$
$x \times 10^{- 2} \times 1000 \times 1 = 12.5 \times 0.03 \times 5 \times 10$
$x = 1.875 mole$

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meq of Fe+2= meq ofKMnO4 x×10−2×1000×1=12.5×0.03×5×10 x=1.875 mole

$meq of F e^{+ 2} = meq of K M n O_{4}$
$x \times 10^{- 2} \times 1000 \times 1 = 12.5 \times 0.03 \times 5 \times 10$
$x = 1.875 mole$