Limiting Reactants Mini Lab

Limiting Reactants Mini Lab

Results:
Data Table of Quantitative Observations:
riddle resistanceMoles of NaI reactedMoles of Pb(NO3)2 ReactedMass of Empty electron tubeMass of Tube with PbI2 ppt.Mass of PbI2 ppt.
11.0 x10 -31.5 10 -47.58g7.65g.07g
21.0 x10 -32.5 10 -48.33g8.43g.10g
31.0 x10 -35.0 10 -48.19g8.43g.24g
41.0 x10 -37.5 10 -46.72g6.97g.25g
51.0 x10 -31.0 10 -38.98g9.12g.23g

Calculations:
Moles of NaI reacted (for completely tubes)
Volume used/1 x 1L/1000mL x groyne/1L
2.0mL/1 x 1L/1000mL x .5/1L = 1.0 x10-3

Moles Pb(NO3)2 Reacted
Volume used/1 x 1L/1000mL x mol/1L
block out Tube 1
.3mL/1 x 1L/1000mL x .5/1L = 1.5 x10-4
adjudicate Tube 2
.5mL/1 x 1L/1000mL x .5/1L =2.5 x10-4
trial run Tube 3
1.0mL/1 x 1L/1000mL x .5/1L = 5.0 x10-4
show Tube 4
1.5mL/1 x 1L/1000mL x .5/1L = 7.5 x10-4
Test Tube 5
2.0mL/1 x 1L/1000mL x .5/1L = 1.0 x10-3

Determine LR
Test Tube One
(.001mol NaI / 1.5 x 10-4mol Pb(NO3)2) = 6.66
(1 / 1) = 1
6.66 > 1
Pb(NO3)2 is the Limiting Reactant
Test Tube Two
(.001mol NaI / 2.5 x 10-4mol Pb(NO3)2) = 4
(2 / 1) = 2
4 > 2
Pb(NO3)2 is the Limiting Reactant
Test Tube Three
(.



001mol NaI / 5 x 10-4mol Pb(NO3)2) = 2
(2 / 1) = 2
2 = 2
Test Tube number Three is the hire stoich ratio
Test Tube Four
(.001mol NaI / 7.5 x 10-4mol Pb(NO3)2) = 2
(2 / 1) = 2
1.33 < 2
NaI is the Limiting Reactant
Test Tube Five
(.001mol NaI / .001mol Pb(NO3)2) = 1
(2 / 1) = 2
1 < 2
NaI is the Limiting Reactant



Determining the Theoretical Yield
Limiting Reactant  PbI2
Test Tube One
(1.5x10-4mol Pb(NO3)2 / 1)(1PbI2 / 1Pb(NO3)2)(460.99g / 1mol PbI2) = .0691g PbI2
Test Tube two
(2.5x10-4mol Pb(NO3)2 / 1)(1PbI2 / 1Pb(NO3)2)(460.99g / 1mol PbI2) = .115g PbI2
Test Tube Three
(5x10-4mol Pb(NO3)2 / 1)(1PbI2 / 1Pb(NO3)2)(460.99g / 1mol PbI2) = .23g PbI2
Test Tube Four
(.001mol NaI / 1)(1Pb(NO3)2 / 2NaI)(460.99g / 1mol PbI2) = .23g PbI2...If you want to get a full essay, ordinate it on our website: Ordercustompaper.com



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