Question

The ratio of number of moles of KMnO4 and K2Cr2O7 required to oxidise 0.1 mole Sn+2 to Sn+4 in acidic medium

Answer 1

2 KMnO4 + 5 Sn{2+} + 16 H{+} → 5 Sn{4+} + 2 Mn{2+} + 8 H2O + 2 K{+}  

(0.1 mol Sn{2+}) x (2 mol KMnO4 / 5 mol Sn{2+}) = 0.04 mol KMnO4  

K2Cr2O7 + 3 Sn{2+} + 14 H{+} → 3 Sn{4+} + 2 Cr{3+} + 7 H2O + 2 K{+}  

(0.1 mol Sn{2+}) x (1 mol K2Cr2O7 / 3 mol Sn{2+}) = 0.03333 mol K2CrO7  

(0.04 mol KMnO4) / (0.03333 mol K2CrO7) = 0.12 / 0.1 = 1.2 = 6/5  

So the ratio required to oxidize is 6:5

Answer 2

Answer:

Explanation:

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

= 6:5