125.0g each of two solids, lead (II) nitrate and pottassiuk iodide are dissolved in the same beaker. A reaction occurs that produces a precipitate.
How many grams of precipitate are produced in the reaction?
This is a limiting reagent (LR) problem. You know that because BOTH reactants are listed.
Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
mols Pb(NO3)2 = grams/molar mass = ?
mols KI = grams/molar mass = ?
Using the coefficients, convert mols KI to mols of PbI2.
Do the same and convert mols Pb(NO)3 to mols KI.
It is likely that the two values for mols PbI2 will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for the lower value is called the LR.
Using the smaller value, convert to grams PbI2.
grams = mols x molar mass = ?
Post your work if you get stuck.
Using the coefficients, convert mols KI to mols of PbI2.
Do the same and convert mols Pb(NO)3 to mols KI.
It is likely that the two values for mols PbI2 will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for the lower value is called the LR.
Tad confused by your instructions
OK. How many mols Pb(NO3)2 do you have? How many mols kI do you have?
What about the converting using coefficients do you not understand?
1 Pb(NO3) and 2 mols of Kl?
That isn't right and you didn't follow what I wrote at the beginning.
mols Pb(NO3)2 = grams/molar mass = ?
mols KI = grams/molar mass = ?
Mols of Pb(No3) 2 = .377?
Mols of kl = .753?
That's a good start.
mols Pb(NO3)2 = grams/molar mass = 125/331.2 = 0.377
mols KI = 125/166 = 0.753
Now the next step is to convert mols of each to mols of the product. You say the directions I gave are confusing. What do you not understand about them?
You start with 0.377 mols Pb(NO3)2 and convert to mols KI produced. You use a factor made up of the coefficients in the balanced equation; i.e.,
0.377 x factor = mols PbI2.
0.753 x factor = mols PbI2.