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equation of potassium iodide and lead nitrate

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Question:What happens when you mix these two chemicals together. I did a science lab and forgot to write these down so I'm kinda screwed ^_^

Answers:Pb(NO3)2(aq)+2 KI(aq)---> PbI2(s)+2 KNO3(aq) PbI2 is a pretty yellow precipitate that is used in making yellow oil paint.


Answers:moles KI = 500 g ( 1 mol / 166.0028 g)=3.01 the balanced equation is 2 KI + Pb(CH3COO)2 = PbI2 + 2 CH3COOK the ratio between KI and PbI2 is 2 : 1 moles PbI2 = 3.01 x 1 / 2 =1.50 mass PbI2 = 1.50 mol x 461.0 g/mol=692 g

Question:In a lab experiment, the reaction of 1.55g of lead(II)nitrate with excess potassium iodide gives 2.00g of lead(II)iodide. What is the percent yield of this reaction?

Answers:Pb(NO3)2 + 2 KI = PbI2 + 2 KNO2 moles Pb(NO3)2 = 1.55 g /331.2 g/mol= 0.00468 = moles PbI2 theoretical amount PbI2 = 0.00468 mol x 461.05 g/mol=2.16 g % = 2.00 x 100/ 2.16 =92.6

Question:Please include balanced equation and justification. Thanks. God Bless.

Answers:The equation is: Pb(NO3)2(aq) + MgI2(aq) PbI2(s) + Mg(NO3)2 (aq) The Pb2+ ions combine with the I- ions to form PbI2 which is insoluble in water , so a precipitate is formed. The Mg(NO3)2 remains in solution as Mg2+ ions anf NO3- ions

From Youtube

Lead Nitrate and Potassium Iodide :DO NOT ATTEMPT THIS DEMONSTRATION AT HOME. Transparent (and mostly colorless) solutions of lead nitrate and potassium iodide are mixed to produce the bright yellow precipitate lead iodide and aqueous potassium nitrate.

reaction of LEAD NITRATE & POTASSIUM IODIDE solutions :When a solution of lead nitrate is poored into a solution of Potassium iodide a yellow precipitate of lead iodide is formed.