observation & Data a) mass of lead (ll) nitrate = 3.88g b) mass of potassium iox

Question

observation & Data a) mass of lead (ll) nitrate = 3.88g b) mass of potassium ioxdide = 4.09g c) mass of filter paper= 1.05g d) mass of filtered paper+precipitate = 5.15g e) mass of precipitate = 4.10g observations= when water was added the product turned a deepyellow. Looks like dry paint, very flakey
Questions to be answered: 1) Write the balanced chemical equation for the doublereplacement reaction. 2) Determine which of the reactants is the limiting factor andwhich is the excess. 3) Using mass-mass calculations, find the theoretical mass ofPbl2 precipitate that should be produced when3.31 grams of Pb(NO3)2 reacts completely. 4) Find the experimental mass of the Pbl2 precipitate. (d)-(c) 5) Determine the percentage error in your experiment. observation & Data a) mass of lead (ll) nitrate = 3.88g b) mass of potassium ioxdide = 4.09g c) mass of filter paper= 1.05g d) mass of filtered paper+precipitate = 5.15g e) mass of precipitate = 4.10g observations= when water was added the product turned a deepyellow. Looks like dry paint, very flakey
Questions to be answered: 1) Write the balanced chemical equation for the doublereplacement reaction. 2) Determine which of the reactants is the limiting factor andwhich is the excess. 3) Using mass-mass calculations, find the theoretical mass ofPbl2 precipitate that should be produced when3.31 grams of Pb(NO3)2 reacts completely. 4) Find the experimental mass of the Pbl2 precipitate. (d)-(c) 5) Determine the percentage error in your experiment. 5) Determine the percentage error in your experiment.

Explanation / Answer

a) The balanced chemical equation of the given reactionis                   Pb(NO3)2 (aq) + 2KI(aq)----------> PbI2(aq) + 2KNO3(aq) b) The number of moles of the lead nitratepresent   = weight / M.Wt                                                                               = 3.88g/331.2g/mole                                                                               = 0.01171 mole The number of moles of the potassiumiodide          =4.09g / 166g/mole                                                                               = 0.0246 mole According to reaction 1 mole of the lead nitrate reactedwith 2 moles of KI.But we have              0.0246 mole KI / 0.01171 mole lead nitrate = 2.1 mole KI / 1mole Pb(NO3)2 Thus there is no suffiecient moles of the lead nitrateto consume all the KI.Thus lead nitrat islimiting reactant. c) The n umer of moles of Pb(NO3)2present = 3.31g/331.2g/mole                                                                        = 0.01 mole Thue the theoridical yield obtained in thereaction                                             =0.01 mole Pb(NO3)2 * (1mole PbI2 / 1 mole Pb(NO3)2)(166g / 1 molePbI2)                                             =1.66g of PbI2 obtained d) Experimental weight of the lead iodide  = mass of filtered paper+precipitate - mass of filterpaper                                                                   = 5.15g - 1.05g                                                                   = 4.10g e) percent of error    =(  4.10 - 4.09 / 4.10) *100)                                 = 0.24%                                                                        = 0.01 mole Thue the theoridical yield obtained in thereaction                                             =0.01 mole Pb(NO3)2 * (1mole PbI2 / 1 mole Pb(NO3)2)(166g / 1 molePbI2)                                             =1.66g of PbI2 obtained d) Experimental weight of the lead iodide  = mass of filtered paper+precipitate - mass of filterpaper                                                                   = 5.15g - 1.05g                                                                   = 4.10g e) percent of error    =(  4.10 - 4.09 / 4.10) *100)                                 = 0.24%

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