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Limiting Reagent Lab. Purpose To determine which compound was the limiting reagent in a reaction between lead (II) nitrate and potassium iodide (referred from the lab sheet) To determine the theoretical value of lead (II) iodide

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Introduction

Limiting Reagent Lab Purpose * To determine which compound was the limiting reagent in a reaction between lead (II) nitrate and potassium iodide (referred from the lab sheet) * To determine the theoretical value of lead (II) iodide and potassium nitrate * To determine the percent yield of lead (II) iodide (referred from the lab sheet) Materials Lead (II) nitrate, potassium iodide, distilled water, 250mL beaker, graduated cylinder, watch glass, triple beam balance, stirring rod, filter paper, funnel, laboratory oven Method In order to determine the limiting reagent and the theoretical and percent yield of products, the following process should be done. * Find a balanced equation of the chemical reaction * Find the limiting reagent using the molar ratio and moles of reactants * Find the theoretical value of a product using the molar ratio between limiting reagent and the product, mole of limiting reagent, and molar mass of the product * Find the percent error using the theoretical and experimental value which is gained by measuring the mass Procedure * Refer to lab sheet * Revision: 1.1 grams of lead (II) ...read more.

Middle

nitrate. Molar mass = mmchemical formula Mole = n chemical formula Mass = mchemical formula Mass mKI = (21.60� 0.05 g) - (20.80� 0.05 g) = 0.80� (Refer to Table 1) Calculation of uncertainty: 0.05 g + 0.05 g = 0.10g mKI = 0.80 � 0.10g (Table 2) mPb(NO 3 ) 2 = 1.10 � 0.10g (Table 2) Molar mass mmKI= 39.10g/mol + 126.90 g/mol = 166.00 g/mol mmPb(NO 3 ) 2 = 331.21 g/mol Mole n = molar mass Mole ratio KI : Pb(NO3)2 = 2 : 1 mole � molar ratio KI: 0.0048 � 1 = 0.0024 2 Pb(NO3)2: 0.00332 � 0.0024 � 0.00332 KI Pb(NO3)2 1 = 0.003 32 1 Potassium iodide is the limiting reagent and lead nitrate is the excess reagent. 3. Theoretical value and percent yield T = theoretical mass E = experimental mass Lead (II) iodide nKI = 0.004819mol mPbI = 460.99 g/mol 2 KI: PbI2 = 2 : 1 1 mPbI = 2 2 � 460.99 g/mol � 0.004819 mol = 1.111 g = T (Table 3) mPbI = 1.41 � 0.10g = E (Table 3) ...read more.

Conclusion

The mass of can be fou KNO3 ii) 2nd method: First, add enough amount of KI to the filtrate so that left over Pb(NO3 )2 is all used up. Now the limiting reagent is not Pb(NO3)2 KI , and therefore the amount of produced is dependent on the amount of . PbI2 Pb(NO3)2 Remove produced using filter paper and heat it in laboratory oven to obtain PbI2 the mass. From the mass of , calculate the amount of in the PbI2Pb(NO3 )2 filtrate using the molar ratio. Both methods may not produce accurate results since they involve filtering. Filtering may cause errors because all liquid cannot perfectly pass through the filter paper. However, both methods theoretically allow one to obtain the mass of KNO3 or excess reagent left. Evaluation 1. Limitations i) All liquid products cannot completely pass through the filter paper. Some liquid is always left on the paper with the solid. ii) Some precipitate sticks to beakers and stirring rods. 2. Effects on result i) Gives less accurate mass of the liquid product. ii) Gives less accurate mass of the precipitate. 3. Improvement i) Find a different way to filter solid form liquid. ii) Try using different tools to clean out the precipitate stuck on lab apparatus. ?? ?? ?? ?? ...read more.

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