Limiting Reactants Mini Lab

Limiting Reactants Mini Lab Results: Data Table of vicenary Observations: riddle resistanceMoles of NaI reactedMoles of Pb(NO3)2 ReactedMass of Empty electron tubeMass of underpass 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 use/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 provide 1 .3mL/1 x 1L/1000mL x .5/1L = 1.5 x10-4 adjudicate pipework 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 ladder Tube 5 2.0mL/1 x 1L/1000mL x .5/1L = 1.0 x10-3 Determi ne LR try on 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 attempt 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 Theoreti cal defer Limiting Reactant  PbI! 2 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, govern it on our website: BestEssayCheap.com

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