Mass Relationships in Chemical Reactions Essay Example for Free

corporation Relationships in Chemical Reactions EssayAimThe aim of this experiment is to battle array that a reply doesnt have always 100% yield by reacting NaHCO3 and HCl and determining the amount of the products to await actual yield.IntroductionA chemical reaction will be quantitative if one of the reactants is all in all consumed. In this experiment sodium bicarbonate and hydrochloric acid start a reaction. The formula of this reaction is below.NaHCO3 + HCl NaCl + urine + CO2ObservationsIn this experiment, sodium bicarbonate is put in an evaporating dish and whatever amount of HCl is added in the dish and the reaction started. Bubbles are formed and CO2 gas is produced and the reaction started to make sound. There was in like manner water vapor formed. White NaHCO3 started to turn into a colorless liquid after adding HCl. As the reaction takes place water is started to form. NaCl was dissolved in water, so salty water is heated to obtain NaCl. As the liquid is heat ed it turned into a yellowish color for a few seconds. Then it started scintillating and water vapor is formed.Raw DataTrial Mass of Dish+NaHCO3+ lid+- 0.1 (g)Mass of NaCl+ pee+Dish+Lid+- 0.1 (g)Mass of NaCl+Dish+Lid+- 0.1 (g)164.14 g.72.16 g.63.28 g.265.14 g.72.95 g.63.91g.Mass of Evaporating Dish + Lid 62.14 +-0.1 gProcessed DataTrial 164.14 62.14 = 2 g NaHCO372.16 62.14 = 10.02 g NaCl + H2O63.28 62.14 = 1.14 g NaClTrial 265.14 62.14 = 3 g NaHCO372.95 62.14 = 10.81 g NaCl + H2O63.91 62.14 = 2.07 g NaClTrial Mass of NaHCO3 (g)Mass of NaCl + H2O (g)Mass of NaCl (g)12 g10.02 g1.14 g23 g10.81 g1.77gCalculationsNa 14.01 g/ jetty, H 1.01 g/bulwark, Cl 35.45 g/mol, O 16 g/mol, C 12.01 g/molNaCl= 49.46 g/molH2O= 18.02 g/molNaHCO3 75.03 g/molMole number of NaHCO3 = mole number of NaClTrial 12 / 73.03 = 0.0274 mol NaHCO31.14 / 49.46 = 0.0230 mol NaCl suppositional Yield 0.0274 mol NaClPercent Yield 0.0230 / 0.0274 = 0.8394 x 100 = 83.94%Trial 23 / 73.03 = 0.0411 mol NaHCO31.77 / 49. 46 = 0.0358 mol NaClTheoretical Yield 0.0411 mol NaClPercent Yield 0.0358 / 0.0411 = 0.8710 x 100 = 87.10%ConclusionThe results are 83.94% for trial 1 and 87.10% for trial 2. Trial 2 is more accurate. The accepted value is 100%. The role errors are 16.06% for trial 1 and 12.90% for trial 2. The uncertainties are too small to calculate on the results. random errors presented in this experiment. All the errors were done by human beings. There werent any errors due to a dishonor of a machine or the procedure.EvaluationWhen salty water is heated on the first trial, the sum total started to spill around, because the substance is heated with high amount of heat and faster than it should be. As a result, some of the NaCl which stuck on the lid and spilled around was lost, so the result of the first experiment is not accurate. Other reasons that changed the results whitethorn be all NaHCO3 may not be dissolved. Too much HCl may be added on the dish. There may be still water molecules lef t on the salt after heating. To overprotect more accurate results, the experiment should be done more slowly than this experiment. Especially the heating butt should be done slowly, so the evaporation can be observed more carefully.