[konkani1kaiser]

Hi,

i have a summer assignment for my AP Chemistry class, and im almost done except for the last problem. I cant find help online and i dont have a textbook, could someone please help me throught this problem...
6) When solutions of potassium iodide and lead (II) nitrate are combined, the products are potassium nitrate and lead (II) iodide.
a. Write a balanced equation for this reaction, including (aq) and (s).
  this was pretty easy, 2KI (aq) + Pb(NO3)2 (aq) ------> 2KNO3 (aq) + PbI2 (s)

after the equation, im not sure where to begin, i can convert M to mols but the questions eludes me. how does one find "precipitate" what is precipitate?

b. Calculate the mass of precipitate produced when 50.0mL of 0.45M potassium iodide solution and 75mL of 0.55M lead (II) nitrate solution are mixed.

c. Calculate the volume of 0.50M potassium iodide required to react completely with 50.0mL of 0.50M lead (II) nitrate.

[sdekivit]

for stoichiometry, you need to use mols. So convert concentration KI and Pb(NO3)2 to mols and use the coefficient to calculate how much PbI can be formed in theory.

c) same as question b as you must know that you need double the amount of mols KI.

[konkani1kaiser]

ok thanks for the start, after converting to moles, .0225mol KI and .0375mol Pb(NO3)2,  i determined that the KI is the limiting reagent, and the theoretical yield is .01125mol of PbI2, so now, should i just forget about the KNO3, and where should i go from the theoretical yield, cuz i still dont know what precipitate means.

[konkani1kaiser]

ok well, here is what i got so far:

a) 2KI + Pb(NO3)2 ---> 2KNO3 + PbI2

b) .05L x .45mol/1L = .0225mol KI, .075L x .5mol/1L = .0375mol Pb(NO3)2
   
    .0225 mol KI x (1 mol PbI2 / 2 mol KI) = .01125 mol PbI2
    .0375 mol Pb(NO3)2 x (1 mol PbI2 / 1 mol Pb(NO3)2) = .0375 mol PbI2
    KI is the limiting reagent, giving a theoretical yield of .01125 mol PbI2.
    from here how i calculate the mass of the precipitate, i do not know

c) .05L x .5mol/1L = .025 mol lead (II) nitrate
   .025 mol Pb(NO3)2 x (2 mol KI / 1 mol Pb(NO3)2) = .05 mol KI
   .05 mol KI x (1 L / .5mol) = .100L or 100ml
   is that right?

[xiankai]

precipitate = salt produced from a reaction that is visible. in this case, it is the product in the (s) state not (aq) state.

[sdekivit]

ok well, here is what i got so far:

a) 2KI + Pb(NO3)2 ---> 2KNO3 + PbI2

b) .05L x .45mol/1L = .0225mol KI, .075L x .5mol/1L = .0375mol Pb(NO3)2
   
    .0225 mol KI x (1 mol PbI2 / 2 mol KI) = .01125 mol PbI2
    .0375 mol Pb(NO3)2 x (1 mol PbI2 / 1 mol Pb(NO3)2) = .0375 mol PbI2
    KI is the limiting reagent, giving a theoretical yield of .01125 mol PbI2.
    from here how i calculate the mass of the precipitate, i do not know

c) .05L x .5mol/1L = .025 mol lead (II) nitrate
   .025 mol Pb(NO3)2 x (2 mol KI / 1 mol Pb(NO3)2) = .05 mol KI
   .05 mol KI x (1 L / .5mol) = .100L or 100ml
   is that right?

you know the amount of mols of the precipitate (PbI) and thus you can use the molar mass to convert to mass.