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Unformatted text preview: H20 and 5ml of ethanol, and re-filtered the crystals. Data: 5.002 of Fe(NH 4 ) 2 (SO 4 ) 2 * 6H 2 O To 2.638 of K 3 [Fe(C 2 O 4 ) 3 ] * 3 H 2 O Calculations: 5.002g / 392.1g/mol = 0.012757 mol Fe(NH 4 ) 2 (SO 4 ) 2 * 6H 2 O 2.638g / 491.1g/mol = 0.005372 mol K 3 [Fe(C 2 O 4 ) 3 ] * 3 H 2 O 0.012757mol 0.005372 mol = 0.007385 mol Conclusion: In this lab we reacted Fe(NH 4 ) 2 (SO 4 ) 2 * 6H 2 O the with excess reactants, and we maintained the number of moles of the product from one reaction to the next. We ended up losing 0.007385 moles of the product to inefficient lab techniques. These would include the filter, the precipitate not completely falling out of solution, the inconsistent heat of reaction, and even an over boiling (bumping)....
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This note was uploaded on 04/30/2008 for the course CHEMISTRY 112 taught by Professor Whelan during the Spring '08 term at UMass (Amherst).
- Spring '08