Percent Yeild Lab

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Percent Yield Lab |
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4/20/2012|
Mrs.Sardella
Per4
Matt , Kait
Mrs.Sardella
Per4
Matt , Kait

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Introduction
*Limiting Reactant: A reactant that is completely consumed during a chemical reaction, limiting the amount of product that is produced. *Excess Reactant: A reactant that remains after a reaction is over. *Theoretical Yield: The amount of product that is predicted by stoichiometric calculations *Actual Yield: The amount of product that is recovered after a reaction is complete. *Percentage Yield: The actual yield of a reaction, expressed as a percentage of the theoretical yield. The equation for calculating the percent yield of a reaction is: Percentage Yield = Actual YieldX100%

Theoretical Yield
Purposes
The purpose of this lab was to determine the limiting and excess reactants for the reaction between potassium iodide and lead (II) nitrate. The second purpose of this lab was to determine the theoretical yield of the reaction. The third purpose of this lab was to determine the percentage yield of the reaction. Pre-Lab Calculations

1. Potassium Iodide (aq) + Lead (II) Nitrate (aq) Potassium Nitrate (aq) + Lead(II) Iodide (s)
2KI(aq) + Pb(NO3)2 (aq ) 2KNO3 (aq) +PbI2 (s) 2.Given/Unkn: 1.25g 1.25g /Ay=? m=? Mm (39.1)+(126.9) (317.2) (39.1) +(14) + (16x3) (207.2)+(126.9x2)

=166g/mol 331.2g/mol 69.4g/mol 461.0g/mol #n (EQUN) 2 mol 1 mol 2mol1mol
M of PbI2 w KI = 1.25g KI X 1mol KI X 1 mol PbI2 X 461.0g PbI2 = 1.735 g of PbI2 LR
166.0g KI 2 mol KI 1 mol PbI2

M of PbI2 W Pb(NO3)2 = 1.25 g Pb(NO3)2 X 1 mol Pb(NO3)2 X 1 mol PbI 2 X 461.0 g PbI2
331.2 g Pb(NO3) 2 1 mol Pb(NO3)2 1 mol PbI2 =1.739 g od PbI2 3.The theoretical yield of the precipitate in the reaction is 1.735 g of potassium iodide. Observation Table
Material| Mass with no substance| Mass with substance| Physical property of Reactants | Beaker A | 66.97g | 68.22g including KI| White Powder| Beaker B| 65.91 g | 67.16 g including Pb(NO3)2| Solid white crystals | Filter paper | 0.56g| 2.16g including PbI2| Yellow Precipitate |

Post Lab Calculations
1. Ay=?
=Mass of the filter paper including substance – mass of filter paper with no substance = mass of PbI2 = 2.16g – 0.56g = 1.6g of PbI2
2. % yield =?
%yield = Actual Mass
Theoretical Mass X 100%
=1.6g
1.735g X 100%
= 92.2 %

Discussion
In this lab potassium iodide and lead (II) nitrate are mixed, and a reaction takes place. The lab begins with two separate beakers. The first beaker held the Potassium Iodide and the second held the Lead nitrate. When deionized water was added and the substances were combined together immediately a precipitate was formed, as well as a bright yellow, cloudy liquid appeared. After the filteration processes the product that our group was left with was a yellow powder consistency. If no sources of error occurred, and the lab took place under perfect conditions the percentage yield of this double displacement reaction should be 100%. However, the percentage yield that my group received 92.2%, which means that somehow throughout the lab process 8%, was lost. There could have been many sources of error. For example, some of them could have been left on the stirring rod, in the beakers, on the funnel or it could have even been spilt. We put a hole in the filter; however we used a new filter and avoided a large error. When measuring the filter paper and its contents the contents of the contents spilled or not placed in the filter paper will produce lower actual yield because there will be less product obtained. All these possible sources of error affected the actual yield thus disturbing the % yield outcome, preventing a perfect 100%. Conclusion...
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