Topics: Solubility, Filter paper, Yield Pages: 5 (1124 words) Published: May 10, 2011
The purpose of this lab was to find the theoretical and experimental percentage yields of the double displacement reaction between the solutions Lead (II) Nitrate (PbNO3) and Potassium Iodide (KI). It is important to obtain amounts of Lead (II) Nitrate and Potassium Iodide as close to 1.44g as possible. This reaction creates Lead (II) Iodide and Potassium Nitrate. The precipitate during this reaction is Lead (II) Iodide. The balanced equation is Pb(NO3)2 + 2KI= PbI2 + 2KNO3. In this lab the Lead (II) Iodide is separated from the solution and collected so it can be weighed for a mass. Potassium Nitrate is separated from the water by boiling the solution, this is resulting in evaporation. This lab is using chemicals and the Bunsen burner so it is required to wear safety glasses throughout the lab. Basic Calculations

Molar Mass
Pb(NO3)2= 331.21 g/mol
2KI= 332 g/mol
PbI2= 460.99 g/mol
2KNO3= 202.22 g/mol
* Approximately 1.44g of PbNO3
* Approximately 1.44g of KI
* Bunsen Burner
* Retort Stand
* Ring Clamps
* Sparker
* Gas hose
* Beaker x3
* Flask x2
* Funnel
* Filter Papers
* Water
* Spatula
* Safety glasses

1) Acquire necessary materials listed above. 1.44g of PbNO3 and 1.44g of KI were acquired. 2) Assemble filtration system. The filter paper was to be folded and inserted into the funnel. Seal the filter paper to the funnel by using water. The funnel was set into the flask. 3) The substances were dissolved in 25 ml of water in beakers. 4) The solutions were thoroughly mixed together and then poured through the funnel. This separated the precipitate from the filter. 5) Separate the precipitate from the filter and place into beaker. 6) Ignite the Bunsen Burner

7) Add 50ml of water to the precipitate. Place upon the Bunsen burner and wait for it to boil. 8) Set up another filtration system. Pour the boiling solution through the filter. The PbI2 can now pass through the filter paper because it is more soluble in its heated state. As the solution cools the crystals form in the purified state. 9) Repeat steps 5-8 to increase the yield.

10) Filter the purified solution.
11) Place the filter with the precipitate within the oven to dry. 12) Weigh the final product.
13) Evaporate the water from the original filtrated solution to find the second substance. 14) Weigh evaporated filtrate.
15) Record results.

Table #1 – Step 1
PbNO3| KI|
Solid| Solid|
White powder| White Powder|
Soluble in room temperature| Soluble in room temperature| 1.44g| 1.44g|

Table #2 – Step 5
PbI2| KNO3|
Liquid| Solid|
Yellow Substance| Whitish in colouration|
Semi clear| Not soluble|
Soluble| |

Table #3 – Step 8
Heated State|
Yellow in colour|

Table #4 – Step 12 and 14
PbI2| KNO3|
Solid| Solid|
Shiny| Dull|
Yellow in colour| White in colour|

Table #5 – Masses of Substances
PbNO3| 1.44g|
KI| 1.44g|
PbI2| 0.16g|
KNO3| 0.68g|

Table #6 – Theoretical Yield
PbI2| 1.99g|
KNO3| 0.879g|

Table #7 – Percent Yield of PbI2
0.16g/1.99gX100%| 8.04%|

Table #8 – Percent Yield of 2KNO
0.68g/0.879gX100%| 77.4%|

A double displacement reaction occurred during this lab between Lead (II) Nitrate and Potassium Iodide. This forms two new substances, Lead (II) Iodide and Potassium Nitrate. The precipitate was PbI2. PbI2 is a yellow substance that is insoluble. In the lab the precipitate was boiled and the substance became soluble. This was caused by heat, it can increase the solubility of substances because when you boil the substance it can be formed into a liquid which makes it soluble. The percent yield of 2KNO was much higher than the percent yield of PbI2. The percent yield of PbI2 was 8.04% while the percent yield of 2KNO was 77.4%. The percent yield...
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