Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO,),(aq) + 2 NH,I(aq) – PbI, (s) + 2 NH,N0; (aq) What volume of a 0.350 M NH4I solution is required to react with 573 mL of a 0.680 M Pb(NO3)2 solution? volume: mL How many moles of PbI, are formed from this reaction? moles: mol PbI2

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter4: Reactions In Aqueous Solution
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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction
РЫ(NO,), (aq) + 2 NH,I(aq) —
PbI, (s) + 2 NH,NO, (aq)
4
What volume of a 0.350 M NH4I solution is required to react with 573 mL of a 0.680 M Pb(NO3)2 solution?
volume:
mL
How many moles of PbI, are formed from this reaction?
moles:
mol PbI2
Transcribed Image Text:Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction РЫ(NO,), (aq) + 2 NH,I(aq) — PbI, (s) + 2 NH,NO, (aq) 4 What volume of a 0.350 M NH4I solution is required to react with 573 mL of a 0.680 M Pb(NO3)2 solution? volume: mL How many moles of PbI, are formed from this reaction? moles: mol PbI2
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