Question Details	Asked By	Asked On
Chemistry help	Phambrick88	12/05/06
How would you verify the limiting reactant of potassium Iodide and Lead (II) Nitrate using a calculation? You can assume there are 20 drops in 1 mL and you used 1 drop of 1.00M solutions of each reactant
Answered By		Answered On
matrix		12/05/06
2 KI + Pb(NO3)2 ---> PbI2 + 2 KNO3 (1/20) x 1.00M = 0.05 mmoles of each of the reagents. You need 2 x 0.05 mmoles = 0.10 mmoles of KI to react with 0.05 mmoles of Pb(NO3)2. Since you only have 0.05 mmoles of KI, at the end of the reaction you will have 0.025 mmoles of Pb(NO3)2 remaining. The KI is the limiting reagent.

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