#### • Class 11 Physics Demo

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# equation of potassium iodide and lead nitrate

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.

Question:

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