Mass Relationships in Chemical Reactions Essay

AimThe aim of this experiment is to show that a reaction doesnt have always 100% yield by reacting NaHCO3 and HCl and find the amount of the products to calculate actual yield.IntroductionA chemical reaction will be quantitative if one of the reactants is completely consumed. In this experiment sodium bicarbonate and hydrochloric acid start a reaction. The formula of this reaction is below.NaHCO3 + HCl NaCl + H2O + CO2ObservationsIn this experiment, sodium bicarbonate is put in an evaporating dish and nigh 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 also water vapor formed. sinlessness 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 modify to obtain NaCl. As the liquid is heat up it turned into a yellowish color for a few seconds. Then it started bubblin g and water vapor is formed. cutting DataTrial aggregate of Dish+NaHCO3+Lid+- 0.1 (g)Mass of NaCl+Water+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/ jetty, 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 NaClTheoretical proceeds 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 true. The accepted cherish is 100%. The percentage 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 through by human beings. There werent any errors due to a flaw of a machine or the procedure.EvaluationWhen salty water is heated on the first trial, the substance 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 chapeau and spilled around was lost, so the result of the first experiment is not accurate. Other reasons that changed the results may be all NaHCO3 may not be dissolved. Too practically HCl may be added on the dish. There may be still water molecules left on the salt after heating. To get more accurate results, the ex periment should be done more slowly than this experiment. Especially the heating process should be done slowly, so the evaporation can be ascertained more carefully.