Ib Chemistry Lab - Products Of a Decomposition Rxn

Name: Laraib Hussain Class: IB1 Lab: Experiment no.5 - PRODUCTS OF A obliteration REACTION Products of a Decomposition Reaction data accrual and info Processing data Collection Quantitative Data - Table of Results: | essay 1| TRIAL 2| Materials| smokestack ( ± 0.01g)| throng ( ± 0.01g)| Clean, Dry melting pot| 14.28| 13.82| crucible + sodium hydrogen carbonate - NaHCO3| 19.28| 18.82| Crucible + Sodium Hydrogen carbonate (after 1st heating)| 17.74| 17.67| Crucible + Sodium Hydrogen Carbonate (after second heating)| 17.67| 17.16| Crucible + Sodium Hydrogen Carbonate (after third heating)| 17.42| 16.95| Crucible + Sodium Hydrogen Carbonate (after quaternate heating)| 17.42| 16.95| qualitative Data - Observations: There were no observations. Data Processing: * differentiate: I chose the results for Trial 2 in order to rank the products of a decomposition reaction of Sodium hydrogen carbonate. computations: 1. experimental Ma ss of Salt:| Calculation(Mass Crucible + Sodium Hydrogen Carbonate (after 4th heating) ) - (Mass Clean, Dry Crucible) 16.95 - 13.82 3.13| Error Calculation 0.01g+ 0.01g= 0.02g0.02/3.13 100 = 0.64 %| ? The observational dope of the salt = 3.13 ± 0.

02 g | From the theory we can accept that NaHCO3 leave break up into eitherNa2O, NaOH orNa2CO3, hence the corresponding realizable equations and calculations be shown below: 1 2NaHCO3(s)?Na2O(s) +CO2(g)+ H2O(l) Calculation| Error Calculation| Moles of : NaHCO35g / 84.01 = 0.06 jetty | ± 0! .01| Moles of : Na2ONaHCO3: Na2O2 : 1? 0.06 × 0.5 = 0.03 mol| ± 0.02| Mass of : Na2OMass = moles Mr0.03 × 61.98 = 1.86 g| | ? The supposititious jalopy of Na2O= 1.86 ±0.02 g / 0.64 %| 2NaHCO3(s)?NaOH(s) +CO2(g) Calculation| Error Calculation| Moles of : NaHCO35g / 84.01 = 0.06 mol | ± 0.01| Moles of : NaOHNaHCO3: NaOH1 : 1? 0.06 ×1 = 0.06 mol| ± 0.02|...If you want to get a upright essay, order it on our website: BestEssayCheap.com

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