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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)
1) what volume of a 0.190 M NH4 i solution is required to react with 223 mL of a 0.200 M Pb(NO3)2 solution?
2) How many moles of Pbl2 are formed for this reaction?
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