Question: Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction: Pb(NO 3 ) 2 (aq) + 2NH 4

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction:

Pb(NO3)2 (aq) + 2NH4I (aq) → PbI2 (s)+ 2NH4NO3 (aq)

What volume of a 0.690 M NH4I solution is required to react with 939 mL of a 0.100 M Pb(NO3)2 solution?

How many moles of PbI2 is formed from this reaction?
 

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