20 mL of K4Fe(CN)6 solution is divided into 2 equal parts. One part is completely reacted with 10 mL of FeCl3 solution to form KFe[Fe(CN)6] & KCl. Another part of K4Fe(CN)6 solution is completely reacted with 20 mL of 0.5 M FeCl2 solution to produce K2Fe[Fe(CN)6] & KCl. Molarity of FeCl3 solution is

Engineering

Chemistry

Law of Chemical Equivalence

Question

20 mL of K₄Fe(CN)₆ solution is divided into 2 equal parts. One part is completely reacted with 10 mL of FeCl₃ solution to form KFe[Fe(CN)₆] & KCl. Another part of K₄Fe(CN)₆ solution is completely reacted with 20 mL of 0.5 M FeCl₂ solution to produce K₂Fe[Fe(CN)₆] & KCl. Molarity of FeCl₃ solution is

$\frac{3}{2} M$

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K₄Fe(CN)₆ + FeCl₃ → KFe [Fe(CN)₆] + 3KCl
n = 3 n = 3
K₄Fe(CN)₆ + FeCl₂ → K₂Fe[Fe(CN)₆] + 2KCl
n = 2 n = 2

${(MV)}_{K_{4} Fe {(CN)}_{6}} = {(MV)}_{{FeCl}_{3}}$
10 × M₁ = M₂ × 10

${(MV)}_{K_{4} Fe {(CN)}_{6}} = {(MV)}_{{FeCl}_{2}}$
10 × M₁ = 0.5 × 20
$M_{{FeCl}_{3}}$ = 1 M