Question

Problems similar to this will be on the scantron-based midterm and final exam If you can successfully complete these you should do well on those tests afnd snosaad a dopdayo Vol. 0.100 M K,CO, solution us o0Z A.2ceol 12oaol Odund uounos CONd N 001 O 19A 1pmal mmole Pb(NO), 00.012 uon o001 oad aded + 1onpoud sseu Jaded Joy ap sseu 514-256 6190 0.256 g 990 323 323 চ 1.2 mmol mmole PbCrO, diaaud po1a d sSeu 1.2nmal 0o0l CALCULATIONS AND QUESTIONS (USE THE SPACE ON THE BACK OF THIS SHEET IF NECESSARY) CrOu+Ph(NO)22Y NO,tPbCro 20% I. In each trial determine which reagent was the limiting reagent? Give a reason, determined by the results of your experiment, which proves your conclusion. tCSONIAS 1.2 ( 4r2) mmol 14 2CNO 1.2 amo 1.4c Po(NDa)a 1-2 mmol 20% 1. 2nmol 32 s9 HOT 2.Calculate the number of millimoles of K,CrO, in 10.55 mL of a 0.120M solution. Report your answer with 3 significant figures (SF). 20% 3.How many milliliters ofa lead(II) nitrate solution (that is 0.1854 M in concentration) is neces- sary to stoichiometrically react with all of the potassium chromate in 45.8 milliliters of a 0092456 M K,CrO, solution? (Work on the back of this page.) Report your answer with 3 SF. 4.If a titration using these solutions started with 25.00 mL of potassium chromate solution (0.5056 M), what volume of 0.2456 M lead(II) nitrate solution would be required to reach the endpoint? What MASS of the products would be created in that reaction? (Work on the back of this page.) Report your answer with 4 SF

please help with numbers 2, 3, 4.

Answer 1

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Number minmolemolity Volume (in m) -23 mmop s 0120 x io.55m 3 Millmole l k,Croy=45.8 mL X o 2456 M millimoles e Pb (N02) - V(mi) X O-1854M at equivatence at end point minimole ek, (r Oy 60.7mL mmdPh(Noz) Vea) = 45.8 mlx o.2456 M O1854 M minimole kroy 05056 M x 25 mL 4. minlmolo af Pb/N), 0-2456 M x V Votameaf Ph(o3 O5056M x 2 5m O 24564 51-4mi ReacHen Pb(No ,Croy Pbro 2KNO3 from abaw ensteniEisclearthat abas PbrafKNo3 mmdaf Pb(No2) an famedLin 2 Ho mmdPbCrDy So Mmol of O 2456 M x 51.47ml =12.64 maud Рьстоy biMp iootfi mass of PbCroy = 12.64x10mot x molar ma 12.64x10mot x 323.2 4.035M mmol of PbNOs mmakNO3 2x12.64 mmd z 25.28 mmet mass KNO325-26x10101 gm/mst 2-SS3 M