Web2KMnO 4+6H ++5H 2O 2→2Mn 2++2K ++8H 2O+5O 2 Thus 2 mol of KMnO 4 require 5 mol of H 2O 2 for decolorization. Therefore for decolourisation of 1 mole of acidified KMnO 4 the moles of H 2O 2 required are 5/2. Solve any question of Redox Reactions with:- Patterns of problems > Was this answer helpful? 0 0 Similar questions Web19 jun. 2024 · 6. A chemist needs 457.8 g of KMnO4 to make a solution. How many moles of KMnO4 is that? 7. How many moles of erythromycin (C37H67N013), a widely used antibiotic, are in 1.00 x 10°g of the
Check: Using the following equation: 5 KNO2 + 2 KMnO4 + 3 …
WebTo calculate molar mass of a chemical compound enter its formula and click 'Compute'. In chemical formula you may use: Any chemical element. Capitalize the first letter in … WebThe number of moles of KMnO 4 reduced by one mole of KI in alkaline medium is: A 1/4 B 2 C 3/2 D 4 Medium Solution Verified by Toppr Correct option is B) Given reaction: KMnO 4+KI→MnO 2+KIO 3 In a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction. Balancing a chemical reaction as: t shirt puma vintage
One mole of acidic KMnO4 reacts with - toppr.com
Web22 mrt. 2024 · Hence, the number of moles of KMnO₄ needed is 0.0536 mole For the mass of KMnO₄ needed, Using the formula, Mass = Number of moles × Molar mass Molar mass of KMnO₄ = 158.034 g/mol Therefore, Mass of KMnO₄ needed = 0.053584 × 158.034 Mass of KMnO₄ needed = 8.468 g Mass of KMnO₄ needed ≅ 8.5 g Web19 aug. 2024 · Then simply multiplying the molarity of the solution by the volume in liters we find the number of moles of MnO4− 15.4mL( 1L 1000mL)(0.0247molMnO − 4 1L) = 3.80 × 104molMnO − 4 The number of moles of oxalic acid, and thus oxalate, present can be calculated from the mole ratio of the reactants in the balanced chemical equation. WebNumber of moles of KMnO 4 used are : A 0.6mol B 1.2mol C 0.4mol D 1mol Hard Solution Verified by Toppr Correct option is A) Fe 2+ as well C 2O 42− is oxidised by MnO 4− in acidic medium. FeC 2O 4⇌Fe 2++C 2O 42− 5Fe 2++MnO 4−→Mn 2++5Fe 3+ 5C 2O 42−+2MnO 4−→Mn 2++10CO 2 5FeC 2O 4≡3MnO 4− ∴1 FeC 2O 4≡ 53MnO 4−=0.6mol philosophy\u0027s 30