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the analyst dissolved 15.32 grams of sodium iodide in 150 mLof water. This solut

ID: 678704 • Letter: T

Question

the analyst dissolved 15.32 grams of sodium iodide in 150 mLof water. This solution of sodium iodide was reacted with 295 mL ofaqueous 0.155 M lead (II) nitrate to produce a precipitate of lead(II) iodide and aqueaous sodium nitriate. The mass of theprecipotate actually generated by the analyst was 8.262grams. Write and balance the equation for this reaction. (Hint wateris a solvent and will not be either a reactant or product in thisreaction) the analyst dissolved 15.32 grams of sodium iodide in 150 mLof water. This solution of sodium iodide was reacted with 295 mL ofaqueous 0.155 M lead (II) nitrate to produce a precipitate of lead(II) iodide and aqueaous sodium nitriate. The mass of theprecipotate actually generated by the analyst was 8.262grams. Write and balance the equation for this reaction. (Hint wateris a solvent and will not be either a reactant or product in thisreaction)

Explanation / Answer

2NaI (aq)+ Pb(NO3)2 (aq)-> PbI2(s) + 2NaNO3 (aq)

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