Question

Plot 1 Mass of precipitate (8) 0.002 0.008 0.01 0.004 0.006 Moles of Pb(NO3)2 Plot 2 Mass of precipitate (g) 0.002 0.008 0.01 0.004 0.006 Moles of KBr Given the plots provided, estimate the following: The number of moles of Pb(NO3)2 that produce a balanced stoichiometry for the reaction. 0.003 0.004 O 0.006 O 0.008

Answer 1

The balanced stoichiometric equation il Pb CNO3), + 2K BY → Pb Byz (S4 + 2 KNO3 Ppt 2+ from the above equation it is clear that pbt and Br- react in 1:2 ratio. to give 1 Pb BY consider option 1 0.003 moles of Pb(NO3) =) According to above cauation, 0.003 moles of PO CONO3)2 reactd with 2x0.003 =0.006 moled of KBY. graph 1 for 0.003 moled Man of the precipitate from of Pb(Nos) is around 19

No. of moled of Pb BV, produced = No. of moled of Pb(NO3)2 reacted = 0.003 moled man of precipitate = moled of Pb BY2 x molar mals of Pb Biz = 0.003 mole x 367.01 g/mol = 1-1 g Eig of graph value. The calculated value matched with the mam of the precipitate (19) at 0.003 molex of graph 1 and at 0.006 moled of grapha. Therefore option 1 il correct. Pb(NO3)2 Produced a balanced stoichioment for the reaction.

ortion 2 Pb(NO3)2 = 0.004 molen 31 KBr = 2x8.004 molen : 0.008 moles man of the precipitate at 0.004 moles of graph 1 is around 1g but man of the precipitate at 0.008 moled of grapha is around o. fg. As the mass of Precinitate is not matching for graph 1 and 2 . This is not correct option. option 3 and 4 0.006 gived KBr as 0.006×2=0.012 moled of K Br are outside the range of grapha Similarly 0.008 given KBr on 0.008X2 = 0.016 moles melen of KBr are outside the range of graph a.