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10.2 g of Pb(NO3)2 are dissolved in 100 mL of water, and added to 250.0 mL...

10.2 g of Pb(NO3)2 are dissolved in 100 mL of water, and added to 250.0 mL of a 0.250 M solution of KI. Determine the mass in grams of PbI2 produced, and find the number of moles of the excess reagent.

I could only find the net ionic equation of: Pb 2+ (aq) + 2 I- (aq) -> PbI2 (s)

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Answer #1

First calculate mole of both reactant and find limiting reagent.

Mass of PbI2 = 14.2 grams

Moles of excess reagent = 0.0625 mole

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