observation & Data a) mass of lead (ll) nitrate = 3.88g b) mass of potassium iox
ID: 678521 • Letter: O
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 flakeyQuestions 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%Related Questions
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