Question

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

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


Pb(NO₃)₂(aq)+2 NH₄ I(aq) ⟶ PbI₂(s)+2 NH₄ NO₃(aq)


What volume of a 0.690 M NH₄ I solution is required to react with 707 mL of a 0.360 M Pb(NO₃)₂ solution?


How many moles of PbI₂ are formed from this reaction?




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