Ratio of number of moles of KMnO4 and K2Cr2O7 required to oxidise 0.1 mole of Sn2+ to Sn4+ in acidic medium.

Open in App

Solution

2 KMnO₄ + 5 Sn²⁺ + 16 H⁺ → 5 Sn⁴⁺ + 2 Mn²⁺ + 8 H₂O + 2 K⁺
(0.1 mol Sn²⁺) x (2 mol KMnO₄/ 5 mol Sn²⁺) = 0.04 mol KMnO₄

K₂Cr₂O7 + 3 Sn²⁺ + 14 H⁺ → 3 Sn⁴⁺ + 2 Cr³⁺ + 7 H₂O + 2 K⁺
(0.1 mol Sn²⁺) x (1 mol K₂Cr₂O₇ / 3 mol Sn²⁺) = 0.03333 mol K₂Cr₂O₇

(0.04 mol KMnO₄) / (0.03333 mol K₂Cr₂O₇) = 0.12 / 0.1 = 1.2 = $\frac{6}{5}$

Suggest Corrections

Similar questions

K M n O_{4}

K_{2} C r_{2} O_{7}

S n^{2 +}

S n^{+ 4}

K M n O_{4}

K_{2} C r_{2} O_{7}

0.1 m o l S n^{2 +}

S n^{+ 4}

K M n O_{4}

K_{2} {C r}_{2} O_{7}

{S n}^{2 +}

{S n}^{+ 4}

{K M n O}_{4}

{K M n O}_{4}

Join BYJU'S Learning Program