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13 Dec 2019

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 0.590M NH4I solution is required to react with 619 mL of a 0.300M Pb(NO3)2 solution?

How many moles of PbI2 are formed from this reaction?

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Jarrod Robel
Jarrod RobelLv2
17 Dec 2019

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