Limiting Reactants Lab
LMounika Alluri
Chemistry HL
Block D
19/9/12
LAB REPORT #1
LIMITING REAGENTS INVESTIGATION
Aim: To determine the limiting reagent and percent yield of the reaction between potassium iodide with lead (II) nitrate solution.
Apparatus required: Safety glasses, funnel stands, watch glass, oven, electronic balance, wash bottle with distilled water, test tubes, 10.0mL 0.50M lead (II) nitrate, 10.0mL 0.30M of potassium iodide solution, two 100.0 mL beakers, funnel, filter paper.
Reaction Involved in Experiment: 2KI + Pb(NO3)2 PbI2 + 2KNO3
DATA TABLE:
SubstanceUsed | Volume (dm3)± 0.0005 | Concentration (M) | Mass (g)±0.001 | Kl | 0.01 dm3 | 0.30 | --- | Pb(NO3)2 | 0.01 dm3 | 0.50 | --- | PbI2 (Residue) | --- | --- | 0.625 g |
Part 1 (To determine the limiting reagent): * Given information: * In this experiment, 0.003mol reacts with 0.0005mol * Concentration of Kl solution = 0.30M * Volume of Kl solution = 0.01 litre ± 0.0005 litre * Concentration of Pb(NO3)2 solution = 0.50M * Volume of Pb(NO3)2 solution = 0.01 litre ± 0.0005 litre
Formula: Number of Moles = Concentration x Volume
Therefore,
No of Kl Moles = Concentration x Volume = 0.30 x 0.01 ± 0.0005 = 0.003mol ± 0.000015mol
No. of Pb(NO3)2 Moles = Concentration x Volume = 0.50 x 0.01 ± 0.0005 = 0.005mol ± 0.000025mol Also, The Theoretical Mole Ratio would be: 21=2 The Experimental Mole Ratio would be: 0.0030.005=0.6 Therefore Since, the ratio theoretical mole ratio > the experimental mole ratio 2 > 0.6 KI is the limiting reagent.
Part 2: (To determine the Percentage Yield) * Given Information: * 0.003mol ± 0.000015mol of KI is reacted To find the number of moles of Pbl2 that is formed with the given
Chemistry HL
Block D
19/9/12
LAB REPORT #1
LIMITING REAGENTS INVESTIGATION
Aim: To determine the limiting reagent and percent yield of the reaction between potassium iodide with lead (II) nitrate solution.
Apparatus required: Safety glasses, funnel stands, watch glass, oven, electronic balance, wash bottle with distilled water, test tubes, 10.0mL 0.50M lead (II) nitrate, 10.0mL 0.30M of potassium iodide solution, two 100.0 mL beakers, funnel, filter paper.
Reaction Involved in Experiment: 2KI + Pb(NO3)2 PbI2 + 2KNO3
DATA TABLE:
SubstanceUsed | Volume (dm3)± 0.0005 | Concentration (M) | Mass (g)±0.001 | Kl | 0.01 dm3 | 0.30 | --- | Pb(NO3)2 | 0.01 dm3 | 0.50 | --- | PbI2 (Residue) | --- | --- | 0.625 g |
Part 1 (To determine the limiting reagent): * Given information: * In this experiment, 0.003mol reacts with 0.0005mol * Concentration of Kl solution = 0.30M * Volume of Kl solution = 0.01 litre ± 0.0005 litre * Concentration of Pb(NO3)2 solution = 0.50M * Volume of Pb(NO3)2 solution = 0.01 litre ± 0.0005 litre
Formula: Number of Moles = Concentration x Volume
Therefore,
No of Kl Moles = Concentration x Volume = 0.30 x 0.01 ± 0.0005 = 0.003mol ± 0.000015mol
No. of Pb(NO3)2 Moles = Concentration x Volume = 0.50 x 0.01 ± 0.0005 = 0.005mol ± 0.000025mol Also, The Theoretical Mole Ratio would be: 21=2 The Experimental Mole Ratio would be: 0.0030.005=0.6 Therefore Since, the ratio theoretical mole ratio > the experimental mole ratio 2 > 0.6 KI is the limiting reagent.
Part 2: (To determine the Percentage Yield) * Given Information: * 0.003mol ± 0.000015mol of KI is reacted To find the number of moles of Pbl2 that is formed with the given