Question

When aqueous lead (II) nitrate is added to aqueous potassium iodide, the brilliant yellow solid, lead (II) iodide forms (as well as aqueous potassium nitrate).

1. Write out the balanced chemical equation for this precipitation reaction.

2. According to the solubility rules provided in class, would you predict an insoluble precipitate, considering the two reactants added together? Why or why not?
   

If 300.0 mL of a 0.10 M solution of each reactant is added together, how many grams of lead (II) iodide could be produced?
(Hints: This is a limiting reactant problem. Use the volume and molarity provided to calculate the given moles of each reactant. Then, determine how many moles of the product, lead (II) iodide could be produced from the given quantity of each reactant.)

Answer 1

16 Friday June a Pb(NO3)2 (aq) + 2 KI (aq) Pbdq (3) + 2 KNO3(aq) by Pbig is insoluble and thus forom a yellow precipitate e here the limiting reactant is Pb(NO3)2 Moles of Pb(NO3)2 = 300 x 0.10 – 0,03 moles from the balanced equation, I more Pb(NO3)2 geve & male Plot2 0.03 Pb(NO3)2 0.03 mole Pb12 1000 (M. W = 461.01) Mass of Pb2=0.03X 461.01g = 13.838 Saturday