JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

In acidic medium, KMnO_4 acts as an oxidising agent and gets reduced to Mn^ 2+ . The n-factor for KMnO_4 is 5. The n-factors for the given reducing agents are calculated based on the total number of moles of electrons lost per mole of the compound: For FeC_2O_4: Fe^ 2+ Fe^ 3+ + e^ - C_2O_4^ 2- 2CO_2 + 2e^ - Total electrons lost = 3, so n-factor = 3. Equivalents = 1 3 = 3. For Fe_2(C_2O_4)_3: Fe^ 3+ is not oxidised. 3C_2O_4^ 2- 6CO_2 + 6e^ - Total electrons lost = 6, so n-factor = 6. Equivalents = 1 6 = 6. For FeSO_4: Fe^ 2+ Fe^ 3+ + e^ - SO_4^ 2- is not oxidised. Total electrons lost = 1, so n-factor = 1. Equivalents = 1 1 = 1. For Fe_2(SO_4)_3: Neither Fe^ 3+ nor SO_4^ 2- can be oxidised further. n-factor = 0. Equivalents = 0. Total equivalents of the reducing mixture = 3 + 6 + 1 + 0 = 10. Let the number of moles of KMnO_4 required be x. Equivalents of KMnO_4 = x 5. Equating the equivalents of the oxidising and reducing agents: 5x = 10 x = 2 Answer: 2