Question

A 1.04 g sample of KBr is dissolved in water to give 155 mL of solution. This solution is then added to 165 mL of 0.015 M aqueous Pb(NO3)2, in an attempt to remove the toxic lead(II) ions from the solution via precipitation as insoluble PbBr2(s).

The precipitation reaction that occurs is: PB2+(aq) + 2Br(aq) --> PbBr2(s)

What is the maximum mass of product that could be isolated by filtration after the reaction occurs?

Answer 1

M INIMH Salution, The balan M MUNTABILITI A many The balanced equation will be — 2K Branit POGNO3)2 caq, 9 PbBr2 G53 + 2 KNO3 CAP:) here you can see that from 2 mole af kBr, one male uf PbBr2 cso as product furmed given, Weight af KBr= 1.048 malamass of KBr = 119.002 g/mol therefure, noi af moles of KBr = 1:04 =0.0087ml 119.0029 mol loooL now, we have to also find the moles of PEGNO3)2, le, mo. af mole of Pb CNOZIE = Gof5 x 165 ml = 0.0024 mol here you can say that PbGNO3)2 uis less ar kBy us in exces. therefore, from about egn When I mole of Pb(NO3)2 reacts I mole Pb Bra for .. when 0.0024 mole af PbCNOgde react to.ouzuxta WEBBTULITEETTISELICITEETTITTITUTTUUTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTMMITTUNTIMITTITUTTTTTTTTTTTTTTTTTTTM TUTULUBBOTTITUTE 1 mole Pb Bra farmed Oculux Imole Pb Bre' - will be fuimed- umolx 367.01 81 mol is weight of PB Bra CS5= 0.0024 molx 367001 cueight of PbBr2 CSI = 0.8808 g Lanser jie weight of product is 0.880897 ģ | Answer