Investigating Stoichiometry Essay

Quantitative dataThe table shows the megabucks of reactants potassium iodide and lead(II) nitrate, and the concourse of the set up from the response between KI(aq) and Pb(NO3)2(aq). pickle of precipitate from reception between KI(aq) and Pb(NO3)2(aq) toilet of potassium iodide (0.001g)1.701Mass of lead(II) nitrate (0.001g)1.280Mass of permeate melodic theme (0.001g)0.798Mass of precipitate + filter news report (0.001g)2.525Mass of precipitate (0.001g)1.727Qualitative Data1) After pouring the KI(aq) and Pb(NO3)2(aq) antecedent in concert into the beaker, a grump rod was use to stir the solution so as to take shape sure it was mixed properly. However, afterwardsward stirring, when the glass rod was taken go forth, on that point were sm each(prenominal) amounts of precipitate (PbI2(s)) stuck onto the glass rod, and could non be removed.2) While pouring the be inter mixed bag into the filter report card, non all the intermixture was poured into the filter funnel and news report. almost of the mixture was stuck in the beaker even after trying to process it down peeing and scooping it out with the glass rod.3) After filtrating the mixture, it was notice that there were nigh parts of the dribble that was still yellow in colour, with well-nigh PbI2 crystals floating a or so, which meant that or so(prenominal) of the residue (PbI2) passed done the filter news report. make up so, another round of filtration was not carried out.The chemical equation obtained from the answer above2KI(aq) + Pb(NO3)2(aq) > 2KNO3(aq) + PbI2(s) footmark 1) Using stoichiometry, prognosticate the gage 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 pass 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)2Moles of PbI2 = 0.0038646176mol Pb(NO3)2 x= 0.0038646176mol PbI2 Using KI Moles of KI = 1.701g KI x= 0.010246988mol KIMoles 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 1.782g PbI2Step 2) immediately we calculate the veridical mass of PbI2 formed.Mass of filter paper = 0.798gMass of precipitate (PbI2) + filter paper = 2.525gMass of PbI2 produced = 2.525g 0.798g= 1.727gStep 3) Now we calculate the part pay up. per centum translate of PbI2 = PbI2 x ampere-second%= 96.91358025% 96.9% synopsis of ResultsAfter conducting the experiment, it is found that the share rejoin of PbI2 produced was 96.9%, which was rather accurate. However, it was rase than the predicted mass by 3.1%, which could be due to the qualitative results shown above, stochastic 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 apply for stirring, and could not be removed without utilise fingers, which would take away foul the solution. This resulted in the decrease in the actual mass of PbI2 precipitate measured, causing the percent yield to be slightly tear down 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. regular(a) though irrigate was used to lap 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 induct fall 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 birth 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.ConclusionThe 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 ImprovementsIf 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 deaden it down back into the mixture. This would have decrease the difference in the persona 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 repeat 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 do the percent yield closer to 100%.