Moles of K2Cr2O7 used to oxidise 1 mol Fe0.95O to Fe+3 are

Engineering

Chemistry

Law of Chemical Equivalence

Question

Moles of K₂Cr₂O₇ used to oxidise 1 mol Fe_0.95O to Fe⁺³ are

$\frac{0.85}{6}$

$\frac{0.95}{6}$

$\frac{85}{95} \times \frac{1}{6}$

$\frac{85}{95} \times \frac{1}{3}$

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Fe_0.95O
x (0.95) – 2 = 0
x = $\frac{200}{95}$ (Charge on Fe)
(n_eq) $K_{2} {Cr}_{2} O_{7}$ = n_eq (Fe_0.95O)
n × 6 = 1 × $(3 - \frac{200}{95})$ × 0.95
n = $\frac{0.85}{6}$ (mol of K₂Cr₂O₇)