For the reaction: Pb(NO3)2 (aq) + 2 NaI (aq)→ PbI2 (s) + 2 NaNO3 (aq)
a. How many grams of PbI2 will be formed from 25.0 mL of a 2.00 M NaI solution?
b. How many milliliters of a 1.25 M Pb(NO3)2 solution will react with 25.0 mL of a 1.50 M NaI solution?
c. What is the molarity of a 20.0 mL solution of NaI that reacts completely with 60.0 mL of a 0.750 M Pb(NO3)2 solution?
a. Pb(NO3)2 (aq) + 2 NaI (aq)→ PbI2 (s) + 2 NaNO3 (aq)
millimoles NaI = mL x M = 25.0 mLx 2.00 M = 50.0
50.0 mmoles NaI x (1 mol PbI2/2 mols NaI) = 50.0/2 = 25.0 mmoles or 0.025 mols,
Then grams PbI2 = moles PbI2 x molar mass PbI2.
b. millimoles NaI = mL x M = 25.0 mL x 1.50 M = 37.5.
moles Pb(NO3)2 needed = 37.5 mmoles NaI x (1 mol Pb(NO3)2/2 mols NaI) = 37.5/2 = 18.75
Then M = millimoles/mL. You have M and mmoles, Solve for mL?
c.See the solution to part b. All the same except the last step.
Post your work if you get stuck.