Question

can anyone help me finish using the numbers already filled in

3.5 25 Mass of Nal Imo Moles of Nal 149.99 g mol 0.0 2348 Mass of Pb(NO3)2 0.01965 mol Moles of Pb(NO3)2 6.5 x imol Phan a mos 331.12 moes por Molarity of Nal mol/L Molarity of Pb(NO3)2 mol/L Limiting reagent Excess reagent Theoretical yield Mass of filter paper .366 0 Mass of filter paper and solid product 5297 0 0 Mass of solid product (Actual yield) of Pbl2 % yield of Pblz solid

Answer 1

Ane >PbIa Na No2 2 Nat PbNO) Stamdard mole Btoichesmey MassNal= 8.58 Molecub wt amele 1male tmole 144.8 m 3.5 no-mels = 00233maa NaT 14. mct mo mab d NaI valume 'Ltre Mlarity Nar 1 O.0233 mela 1L o.0133Mall L phNo2 S429 Mass moleaulerwt 331. 2 mo homsles Ph N2) S427 331-2 m 0.01976 me Less in mumber Compan to NaI 8 Limiting Agent pbNoz)2 act ao and NaI io excesseagent O,0197mol Mel evityPNo)2selution i L IL Oio976 melsL Reactionn Table Ph NaNor Pb (No) a NaT O01976 am 2xO0233 0,0466no Livmiting aget

giveo Pa-2 1mols Pb(No3) 1mol PbL2 aire 0,01976mob pNo3) 0.01936ob fb . mol PbI2 461-ol 0.01976x461-0I 9.103 gmTheoretical idd Theoretical yrel od PbI 9.loI Herpeper tooliol teltes paper Obserned yield mass 5.297gm 4.931gm 0.36 m 1 obgeneot Yelot Xloa Freld PbI Theeretionl yeld 4.93 g x 1OD 9.109 3m 547 rheoreticul yielol-obemol Thearetical yelod 9.109-.93 7. Enror In cgeiment 9.109 45 87 %
Less in number of moles of reagent act as a limiting agent and used for theoretical yield calculation based on stoichiometric relation.