Equal amounts, unequal shares
Lead nitrate and potassium iodide make a bright yellow precipitate of lead iodide. Pour equal volumes of each at equal strength and it looks like a fair mix. It is not. The equation needs two iodides for every lead, so the iodide runs out halfway. This activity asks you to find that from the numbers, work out the mass of yellow solid, and then see what doubling the iodide does to it.
What you will use
Apparatus
- Beaker
Chemicals
- Lead Nitrate Solution
- Potassium Iodide Solution
The steps
-
Pour lead nitrate solution into the beaker, and choose 0.5 M.
Colourless. Five millimoles of lead ions.
-
Pour potassium iodide solution in, also at 0.5 M.
A yellow solid forms at once and settles: lead iodide.
-
Same volume, same strength. Which reagent ran out first?
Equal moles, but the equation takes them two to one. The one it needs twice as much of runs out.
-
How many grams of lead iodide formed?
Half as many moles of PbI₂ as of KI, times its molar mass of 461 g/mol.
-
Pour another 10 mL of potassium iodide at 0.5 M, so 20 mL in all.
More yellow. The lead that had nothing to react with now does.
-
Now how many grams of lead iodide are there in all?
Twice the limiting reagent, twice the yield. Now the moles match the equation exactly, and neither reagent is left.
What it shows
Equal amounts are only a fair mix when the equation takes them one to one. The numbers in front of the formulas decide what runs out, and that decides how much product you get, however much of the other reagent is waiting.