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The table shows the mass of reactants potassium iodide and lead(II) nitrate, and the mass of the precipitate from the reaction between KI(aq) and Pb(NO3)2(aq).
Mass of precipitate from reaction between KI(aq) and Pb(NO3)2(aq)
Mass of potassium iodide (ï¿½0.001g)
Mass of lead(II) nitrate (ï¿½0.001g)
Mass of filter paper (ï¿½0.001g)
Mass of precipitate + filter paper (ï¿½0.001g)
Mass of precipitate (ï¿½0.001g)
1) After pouring the KI(aq) and Pb(NO3)2(aq) solution together into the beaker, a glass rod was used to stir the solution so as to make sure it was mixed properly. However, after stirring, when the glass rod was taken out, there were small amounts of precipitate (PbI2(s)) stuck onto the glass rod, and could not be removed.
2) While pouring the remaining mixture into the filter paper, not all the mixture was poured into the filter funnel and paper. Some of the mixture was stuck in the beaker even after trying to wash it down water and scooping it out with the glass rod.
3) After filtrating the mixture, it was observed that there were some parts of the filtrate that was still yellow in colour, with some PbI2 crystals floating around, which meant that some of the residue (PbI2) passed through the filter paper. Even so, another round of filtration was not carried out.
The chemical equation obtained from the reaction above:
2KI(aq) + Pb(NO3)2(aq) –> 2KNO3(aq) + PbI2(s)
Step 1) Using stoichiometry, predict the mass of PbI2(s) formed when a solution containing 1.701g of KI(aq) is mixed with a solution containing 1.280g of Pb(NO3)2(aq):
First, the limiting reagent is determined by finding out which reagent produces lesser moles of PbI2.
Using Pb(NO3)2: Moles of Pb(NO3)2 = 1.280g Pb(NO3)2 x
= 0.0038646176mol Pb(NO3)2
Moles of PbI2 = 0.0038646176mol Pb(NO3)2 x
= 0.0038646176mol PbI2
Using KI: Moles of KI = 1.701g KI x
= 0.010246988mol KI
Moles of PbI2 = 0.010246988mol KI x
= 0.005123494 mol PbI2
?Pb(NO3)2 is the limiting reagent.
Second, we predict the mass of PbI2 formed.
Mass of PbI2 = 0.0038646176mol PbI2 x
= 1.781550067g PbI2
ï¿½ 1.782g PbI2
Step 2) Now we calculate the actual mass of PbI2 formed.
Mass of filter paper = 0.798g
Mass of precipitate (PbI2) + filter paper = 2.525g
Mass of PbI2 produced = 2.525g – 0.798g
Step 3) Now we calculate the percent yield.
Percent yield of PbI2 = PbI2 x 100%
Analysis of Results
After conducting the experiment, it is found that the percent yield of PbI2 produced was 96.9%, which was rather accurate. However, it was lower than the predicted mass by 3.1%, which could be due to the qualitative results shown above, random errors and inaccuracy of the experiment.
When stirring the KI(aq) and Pb(NO3)2(aq) solution, some of the PbI2 precipitate was stuck onto the glass rod used for stirring, and could not be removed without using fingers, which would have contaminated the solution. This resulted in the decrease in the actual mass of PbI2 precipitate measured, causing the percent yield to be slightly lower than the predicted yield.
When pouring the mixture into the filter funnel, not all of the mixture was poured into the filter paper as some of it was stuck inside the beaker. Even though water was used to wash some of the mixture stuck in the beaker into the filter paper, not all of the mixture was filtered. The mixture stuck in the beaker and was not filtered would have decreased the percent yield.
Finally, when the mixture was being filtered, some of the PbI2 precipitate passed through the filter paper and went into the filtrate. The filtrate was not filtered again, so some of the PbI2 was not calculated into the final mass of PbI2 produced. This would have decreased the percent yield as well.
All the above would have contributed to the fact that the percent yield was 3.1% lower than the predicted yield.
The results from the experiment showed that the percent yield of PbI2 is 96.9%, which is rather accurate. However, due to random errors and the qualitative results shown above, the percent yield is 3.1% lower than the predicted yield.
Limitations and Improvements
If I could do the experiment again,
–> As some of the precipitate was stuck onto the glass rod and could not be removed by using my fingers, I could have just used a little bit of water to wash it down back into the mixture. This would have decrease the difference in the percentage between the predicted yield and the percentage yield.
–> Even though water was used to wash some of the mixture into the filter funnel, there was still some mixture stuck in the beaker. The process of using water to wash down the mixture could have been repeated over and over until all the mixture is in the filter funnel.
–> After filtrating the mixture once, some of the PbI2 crystals went through the filter paper and into the filtrate in the conical flask. To make sure all the PbI2 precipitate is counted towards the percent yield, the filtrate could have been filtrated again at least 2 more times. This would have increased the mass of PbI2, which would have made the percent yield closer to 100%.