Dry Lab

A dry lab?  Doesn't that just make you think the ingredients or materials are dry? well thats what i thought until Mr. Ludwig explained that this lab was actually very serious and dangerous and we would not actuallly be performing the procedure. In this lab we would simply be following what others have already done, and simply analyzing the conclusion and results.

PROCEDURE TIME!!!!

1. Clean and dry two beakers
2. Record the number of you chemical- Lab #5 ( I had either 4 or 5 I cant really remember which! Sorrys!)
3. Find the mass of lead(II) nitrate- 1.03 grams
4. Find the mass of potassium iodide- 1.33 grams
5. Add about 50mL of DI water to each beaker
6. Stir the beakers until both chemicals have dissolved
7. Combine the contents of both beakers into one beaker
8. Wash any remaining solution into the one beaker
9. Stir the solution
10. Measure the mass of filter paper- 1.39 grams
11. Filter out the precipitate into the filter paper
12. Wash any remaining precipitate into the filter paper 
13. Allow the precipitate and filter paper to dry overnight
14. Find the mass of the precipitate- .95 grams

QUESTION TIME!!!

Question: What is the balanced equation for the chemical reaction that occured in the experiment?

Answer: The balanced equation for this particular chemical reaction is: 2KI + Pb(NO3)2 --> 2KNO3 + PbI2 (Thank goodness for notes! I'm so slow at balancing equations!)

Question: What was the limiting reagent for your experiment?
Answer: The limiting reagent for this experiment was: Lead (II) Nitrate

Question: How much lead (II) Iodide will theoretically be formed?
Answer: Theoretically speaking there should be: 1.43 grams of lead(II) iodide formed from the experiment

Question: How much lead (II) iodide did you get from your experiment?
Answer: However when i performed the experiment there was actually: .96 grams of lead(II) iodide formed from the experiment

Question: What was the Percent yeild for your experiment?
Answer: The percent yield of lead(II) from the experiment was: 67%