Hey. This is my chemestry paper, tell me what you think of it. One of the main reasons I'm posting it here is for a critique, and so I can print it on a different computer. Get ready for some horse shit!
Tim Burns
06.07.06
Per. 02
Making PbI2
Hypothesis: It is possible to synthesize Lead (II) Iodide by mixing Lead (II) Nitrate and Sodium Iodide together, both in an aqueous state.
PbI2 is used to form some pigments and fake gold colors. In order to synthesize this compound one would have to dissolve Pb(NO3)2 in distilled water as well as dissolving NaI in a separate container of distilled water. The use of distilled water is critical because it is easily filtered away and adds nothing that is not easily removable. We decided upon NaI and Pb(NO3)2 because lead (Pb) is more reactive than sodium (Na), that would cause the Na to be replaced by the Pb creating PbI2.
We need to render 11.53g, or .025 moles of PbI2. This will be a final amount needed to turn in. We will need an excess amount of one compound to be assure us that the reaction is complete. We will use NaI as our excess reactant because it is less toxic and more economic. Since it is in excess we won't need a definite measurement on it. NaNO3 is a liquid when mixed, and PbI2 is a crystalline type structure at room temperature that can't be dissolved in water.
Materials:
-Erlenmeyer flasks Three
- Beaker One
- Clean material trays Two
- Filter Paper One-- weigh and mark dry weight.
- Clean funnel One
I used two Erlenmeyer flasks to turn the NaI and Pb(NO3)2 into an aqueous state. I labeled the mixtures before the chemicals where added in order to prevent a mix up. The NaI did not need a set measurement of either water or actual compound mass although with the stoichiometry knowledge and a math problem shown below I learned that in order to have the reaction with a bare minimum amount of reactant I would need 7.5 g of NaI dissolved in distilled water. We will use 15 g of NaI, this will help ensure all of our limiting reactant is being reacted.
Pb(NO3)2 is the limiting reactant, so we need to know the measurements to assure we have an excess of our excess reactant. With the simple math problem below we can tell the minimum amount of Pb(NO3)2 needed to render .025 moles of PbI2. as you can see 8.28 g is the minimum needed.
The amount of water we will use is 30 mL of distilled water to assure that the Pb(NO3)2 is fully dissolved. 92.1 g of Pb(NO3)2 is the maximum amount that can be dissolved in 100 mL of water.
After the two substances are completely dissolved You can pour them into the beaker, after the reaction is formed you can place the filter paper in the funnel and put the funnel in the third clean Erlenmeyer flask. Be sure that the flask is large enough in volume to collect all of the liquid that is filtered out. If you end up with clots of PbI2 use distilled water to aid in flushing it out of the beaker. Let the liquid drain, then wait for the filter to dry with the lead (II) iodide in it. if all is done correctly you should have 11.53 g of pure PbI2.
To dispose of the compounds dump the liquid in a marked liquid waste bucket and the filter paper in the labeled solid waste bucket because they may have traces of lead left in them.
Various masses:
Mass of Pb(NO3)2: 8.28 g
Mass of Filter Paper: .98 g
Mass of filter paper with PbI2: 12.5 g
Mass of PbI2: 11.51 g
% yield: 99.82%
Conclusion:
It does appear that it what I have made is PbI2, there is a 99.8% yield. All of the data has shown that this synthesis is possible and easily done. Some of the errors I had where the Pb(NO3)2 not dissolving easily because of large chunks with less surface area than the pure powder. Also, I had a large excess of NaI and water.
