Lab Report - The Stoichiometry of an Oxidation-Reduction Reaction

Purpose: To find egress of moles of Fe+3 which counterbalance with maven mole of NH3OH+ in order to typeially balance the equality: NH3OH+ + 2Fe+3 ? + 2Fe+2 and to find the lacking(p) product. Procedure: H2C2O4 and H2SO4 were titrated with yard permanganate. The metre of the permanganate was then found because the yard of the H2C2O4 and H2SO4 were already known. hence hydroxylammonium chloride and ferrous sulfate and water was titrated with known potassium permanganate to add up the molarity of the hydroxylammonium chloride and ferric sulfate solution, and that was used to find the tot up of moles of Fe+3 which reacts with one mole of NH3OH+ to find the missing part of the equation. Calculations: 0.103 moles/1000 mL = .00103 moles/10 mL .00103 x (2/5) = .000412/16 mL = .0256 M .0256 moles/1000 mL = 1.95 x 10-4 moles / 7.6 mL (1.95 x 10-4) x 5 = 9.75 x 10-4 moles = moles of Fe+2 .05 moles/1000 mL = .0005 moles/10 mL = moles of hydroxylammonium chloride dimens ion of Fe+2 to NH3OH+ = 2:1 2e- + 2Fe+3 --> 2Fe+2 so transfer of 2 electrons NH3OH+ --> something + 2e- Oxidation number of N in NH3OH+ is -1, thereof the oxidation number for N on the product billet must be +1 because it gains 2 electrons. N2O has an oxidation number of +1 for N, so that would work.

Data: equation 1: NH3OH+ + 2Fe+3 --> something + 2Fe+2 Equation 2: 8H+ + 5Fe+2 + MnO4- --> 5Fe+3 + Mn+2 + 4H2O Equation 3: 6H+ + 2MnO4- + 5H2C2O4 --> 2Mn+2 + 10CO2 + 8H2O Conclusion: Therefore the concluded answer would be: NH3OH+ + 2Fe+3 --> N2O + 2Fe+2 This was obtained by using stoichiometry half reactions the product of that reaction was determined to be N2O. Some s ystematic... ! To make this better for a more commonplace aspect...you could swallow explained what some of the terms were interchangeable I assign some dont know what molarity is, that would have made it better...but I geuss for your class lab report that doesnt matter...oh well, nevermind that suggestion. If you need to get a full essay, order it on our website: OrderEssay.net

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