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Limiting Reagent and Percent Yield
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
* 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.
* 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+16x3)x2 = 331.21m.m.u. Molecular mass of KI: 39.10+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 + 126.90x2 = 460.99m.m.u. Number of moles of PbI2:...
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