terAme ×U10idation States-Chempend * Settings earning.com/ibiscms/mod/ibis/view.
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terAme ×U10idation States-Chempend * Settings earning.com/ibiscms/mod/ibis/view.php?id 4326226 Jumpt -PrintCalolator Period. Table Question 31 of 33 Sapling Learning Map cob Leadl) nitrate and ammonium iodide react to form lead(l) odide and ammonium nitrate according to the Ph(No,.laq) +2NH.I(aq) Pt1iW-2NH,Nolaq) What volume of a 0.150 M NHJ solution is required to react with 987 mL of a 0 240 M P(NOsh solution? nl. How many moles of Pt12 are formed from this reaction? mol Pbl, MacBook Air F3 888 F 3 4. 8 9Explanation / Answer
Pb(NO3)2 + 2NH4I -------------> PbI2(s) + 2NH4NO3
987x0.24 V x 0.150 0 0
For all the lead to be precipitated as PbI2 we need twice the mmoles of NH4I
Thus mmoles of NH4I required = 2 x 987 x 0.24 which sholud be equal to Vx0.150
= V x0.150
thus V= 3158.4 mL
ALTERNATIVELY
use V1M1/n1 = V2M2/n2
part 2)
Pb(NO3)2 + 2NH4I -------------> PbI2(s) + 2NH4NO3
987x0.24 V x 0.150 0 0 initial mmoles
0 0 987x0.24 2x987x0.24 after reaction
Thus mmoles of pbI2 formed = 987x0.24
= 236.88 mmol
Moles of PbI2 formed = 0.236 mol
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