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)ā¶PbI2(s)+2NH4NO3(aq)
What volume of a 0.690 M NH4I
solution is required to react with851 mL of a 0.260 M
Pb(NO3)2 solution?
How many moles of PbI2 are
formed from this reaction?