Question

I’m not sure what I’ve done wrong since the rest of my calculations were correct, so will someone please explain to me how to solve these. I’ve tried everything i can think of and nothing is working.

Answer 1

= 0.02) 6 mol Balanced chemical equation Pb(NO3), (02) + 2 Kallel.) → Pode (>) + 2XN03 (ol.) given weight of Pb(NO3)₂ = 7.158 molar man g Pb(NO3)2 = 331-2 g moins mol of Pb(NO3)2 = given man Molarmas molg Pb(NO3)2 = 7.158 33)-27 mol of kl= 5.oog - 0.0671 mol 74.55 mortº From Balanced reaction I mol g Pb(NO3), react with 2 molku 0.0216 mola Pb(NO3), C mong Pb(NO3), react with 0.0432 0 ku moly ku available or0622 0.0671 but ku used = 0.0432 mol This means Pb(NO3), is limiting reagent and consume completely and reaction proceed according to PbCNO3), so I mol of Pb(NO3)2 give Imol PbCl2 0:0g16 mol of Pb(NO3), give 0:0216 mol Pbila mol q pb Q2 = 0.0215 mo? mang poolz = mol x molar mass = 0.0216 3801 278.99 = 6.00g met Theertical yield of pbcla = 6:01g olo yield = 80.600 % gield = expennental gielt xloo Thaeretical gied experimentel yield xloo 806 = 6olg

experimental yield = 4.8499 since ke present in excess so molke left = 0.0671 mol - 0.0432 mol = 0.0239 mol molg ke lett = 0.0239 mol mass of ku left = 0.0239 mol X74.55g met = 1.782 g mass of Ke = (1.782 g Answer