Limiting Reagent and Percent Yield Aim To determine the limiting reagent between the reaction of lead (II) nitrate and potassium iodide. To determine the percent yield of lead (II) iodide. Date Started: 13/4/12. Finished: 19/4/12. Data collection and processing Measurements: * Amount of distilled water: 75. 0ml ± 0. 5ml. * Mass of watch glass: 31. 65g ± 0. 01g. * Mass of watch glass + potassium iodide: 32. 45g ± 0. 01g. * Mass of potassium iodide: 0. 8g ± 0. 02g. * Mass of watch glass + lead (II) nitrate: 32. 66g ± 0. 01g. Mass of lead (II) nitrate: 1. 01g ± 0. 02g. * Mass of filter paper: 0. 65g ± 0. 01g. * Mass of filter paper + lead (II) iodide precipitate: 1. 65g ± 0. 01g. * Mass of lead (II) iodide precipitate: 1. 00g ± 0. 02g. Observation: * When adding the potassium iodide into the distilled water, no visible reaction occurred. The distilled water kept clear. * When adding the lead (II) nitrate to the solution a visible reaction occurred. The solution turned opaque yellow, the color started spreading along the solution as soon as the lead (II) nitrate was added. After letting the reaction had occur for some minutes. A yellow precipitate was visible at the bottom of the beaker. * When filtering the solution the yellow precipitate was collected in the funnel. * After letting the precipitate dry for some days, it became a yellow powder. Limiting reagent and percent yield of lead (II) iodide: The equation for the reaction: Pb(NO3)2 + 2KI 2K(NO3)+ PbI2 1 mole of Pb(NO3)2 react with 2 moles of KI. Molecular mass of Pb(NO3)2: 207. 19 + (14. 01+16×3)x2 = 331. 21m. m. u. Molecular mass of KI: 39. 0+126. 90 = 166m. m. u. Number of moles of Pb(NO3)2: 1. 01g/331. 21m. m. u. = 0. 00305 mol (rounded). Number of moles of KI: 0. 8g/166m. m. u. = 0. 0048 mol (rounded). There is only 0. 0048 moles of KI, therefore the limiting reagent is KI. Molecular mass of PbI2: 207. 19 + 126. 90×2 = 460. 99m. m. u. Number of moles of PbI2: 1. 00g/460. 99m. m. u. = 0. 0022 mol. When 0. 0048 moles of KI react, moles of PbI2 : 0. 0048/2 = 0. 0024. Theoretical yield: 0. 0024mol x 460. 99m. m. u. = 1. 1g of KI. Therefore the percent yield is: 1g/1. gx100 = 91%. Evaluation Weaknesses and limitations: * Both, potassium iodide and lead (II) nitrate, were not as a powder (chunks), therefore I wasn’t able to measure precise measurements. * I was not able to completely rinse all the lead (II) iodide left on the beaker, some chunks of lead (II) iodide were not possible to rinse. * The amount of the substances used was very small, for example 0. 8g of potassium iodide and 1. 01g of lead (II) nitrate. Using larger amount of the substances the precision will be higher.
In order to remove this weaknesses and limitation one can obtain potassium iodide and lead (II) nitrate as a powder (not affected by humidity). Also one can use larger amounts of substances to increase precision. And finally rinse the beaker for a longer time so that the amount of lead (II) iodide left on the beaker is smaller. Conclusion In conclusion the experiment went very well, there were no major limitation or weaknesses. I was able to measure very precise and accurate readings. I was able to achieve the aim as I was able to calculate the limiting reagent and the percent yield. Limiting reagent: KI. Percent yield: 91%.
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