1.MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ---------> Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)2. Calculate the volume of iron (ii) sulfate into dm^3 (25.0 cm^3= 0.025dm^33.Calculate the number of moles for the potassium permanganate, (volume of Pottasium permangnate used) x (concentration of Potassium permanganate) = (Number of moles of Potassium Permanganate)4. Mole RatioMnO : Fe    1    :  5 5.Repeat for accuracy and calculate the average mass of iron per tablet from multiple measurements

calculate molarity of the standardized potassium permanganate solution in redox titration of iron 2 into iron 3. molecular weight of hydrated ferrous ammonium sulfate is 392.14 g/mol. also use the ration of iron 2 to permanganate ion from equation: 5Fe + Mn04 + 8H gives 5Fe + Mn + 4H2O. VOLUME OF KMnO4 used in titration is 7.65 ml and mass of hydrated ferrous ammonium sulfate is 0.2911 g.

27. Calculate the amount of Fe+2 ions in 500 ml Mohr’s salt solution if 10ml of this solution is titrated with 7.5ml 0.06N KMnO4 solution. (1ml 1N KMnO4= 0.05584 gm of F+2). Ans: 1.256 gm

Calculate the volume (in cm3) of oxygen (measured at STP) produced when 25.0 cm3 of 0.0180 mol dm-3 potassium manganate(VII) reacts with 30.0 cm3 0.0250 cm3 hydrogen peroxide solution in the presence of excess sulfuric acid, according to the equation:2KMnO4(aq) + 5H2O2(aq) + 3H2SO4(aq) → 2MnSO4(aq) + K2SO4(aq) + 5O2(g) + 8H2O(l)

Iron(II) sulfate, is typically the source of iron in tablets prescribed to anemic patients, a condition where a person lacks enough healthy red blood cells. The tablets can be dissolved in dilute sulfuric acid to solvate the Fe2+ ions. To determine the percentage by mass of iron within the tablets, the Fe2+ ions can be oxidised to Fe3+ using potassium permanganate, where MnO4− is reduced to Mn2+.Calculate the percentage by mass of iron in a 500 mg tablet that requires 24.53 cm3 of 0.0100 mol dm−3 KMnO4 to titrate the iron sample. State your answer to 3 significant figures.                M(Fe) = 55.85 g mol−1

Calculate the mass (in g) of potassium permanganate (KMnO4) required to make a 250 mL of a 0.951 M solution. Answer to 3 decimal places.MW(KMnO4) = 158.04 g.mol-1