Chuong Nguyen U84536721 Post Lab Report Experiment 6 go on out B ( ascorbic acid points) 1. (15 points) Prepare a table leaning the caboodle of the starting materials used and the trade of the dried proceeds you collected. mastermind the percent supply for each compound. soften 1| | Part 2| | Material| tote up| Material| Amount| KMnO4| 3.7580 g| [NiCl2(PPh3)2]| 0.6026 g| C5H8O2| 16.6 g| NH4SCN| 0.8024 g| Product| | PPh3| 2.0016 g| Mn(acac)3| 6.119 g| Product| | | | [Ni(NCS)2(PPh3)2]| 1.56 g| % topic= mass of fruit/ mass of reactant x 100% Part I 4 KMnO4 + 12 C5H8O2 4 Mn(C5H8O2)3 + 2 K2O + 7 O2 Theoretical yield 0.027 bulwark KMnO4 x 352.26 g/mol Mn(C5H8O2)3 (1:1 ratio) = 9.51 g Mn(C5H8O2)3 0.166 mol C5H8O2 x 352.26 g/mol Mn(C5H8O2)3 (3:1 ratio) = 19.47 g Mn(C5H8O2)3 Limiting reactant KMnO4 Mn(acac)3 %yield = 6.110 g /9.51 g KMnO4 x 100% = 64.25% Part II [NiCl2(PPh3)2] + 2NH4SCN [Ni(NCS)2(PPh3)2] + 2 NH4Cl Theoretical yield .6026 g /259.54g/mol[NiCl 2(PPh3)2] x 304.80 g/mol [Ni(NCS)2(PPh3)2] (1:1 ratio) = 0.7076 g [Ni(NCS)2(PPh3)2] 0.8024 g/ 76.122g/mol NH4SCN x 304.80 g/mol [Ni(NCS)2(PPh3)2] (2:1 ratio) = 1.606 g Limiting reactant [NiCl2(PPh3)2] [Ni(NCS)2(PPh3)2] % yield = 1.56g / 0.7076 g x 100% = 220.
46 % (Excess triphenylphosphine need to account in chemical reply and production yield) 2. (15 points): Prepare a table listing both the experimental magnetic data you obtained for the three compounds you measured. From the data you accumulated, account the gibibyte force ?g of your samples. Convert your gram susceptibility to hero sandwich suscepti bility, ?M. Compound| L| W| Ro| R| Mn(a! cac)3| 3.0 cm| 0.1744 g| -32| 1714| [NiCl2(PPh3)2]| 3.2 cm| 0.1333g| -32| 220| [Ni(NCS)2(PPh3)2]| 4.0 cm| 0.1414 g| -32| -56| ?g = 1.007 (L)(R-R0)/(109)(m) Mn(acac)3= 1.007(3)(1714- (-32)/ (0.1744 x 109) = 3.024 x10-5 [NiCl2(PPh3)2] = 1.007(3.2)(220 (-32)/ (0.1333 x 109) = 6.091 x 10-6 [Ni(NCS)2(PPh3)2] = 1.007(4)(-56 (-32)/ (0.1414 x...If you want to get a in effect(p) essay, hallow it on our website: OrderCustomPaper.com
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