Mass Relationships in Chemical Reactions Essay

betThe aim of this experiment is to show that a reaction doesnt incur always cytosine% yield by reacting NaHCO3 and HCl and determining the amount of the products to mastermind 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 + carbonic acid gasObservationsIn this experiment, sodium bicarbonate is put in an e vaporating dish and whatever amount of HCl is added in the dish and the reaction started. Bubbles are formed and carbon dioxide gas is produced and the reaction started to make sound. thither was also 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 zesty water is catch fireed to obtain NaCl. As the liquid is heated it saturnine into a yello wish color for a few seconds. Then it started pass off and water vapor is formed.Raw DataTrial the great unwashed of sweetie+NaHCO3+Lid+- 0.1 (g)Mass of NaCl+Water+ sweetie+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/ groynee, H 1.01 g/ mole, 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 admit 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 = 0.0 358 mol NaClTheoretical Yield 0.0411 mol NaClPercent Yield 0.0358 / 0.0411 = 0.8710 x 100 = 87.10%ConclusionThe results are 83.94% for running game 1 and 87.10% for trial 2. Trial 2 is to a greater extent accurate. The accepted value 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 done by human beings. There werent any errors due to a flaw of a utensil or the procedure.EvaluationWhen salty water is heated on the early trial, the substance started to spill almost, 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 may be all NaHCO3 may not be dissolved. Too much(prenominal) HCl may be added on the dish. There may be assuage water molecules left on the salt after heating. To get more accurate results, the experiment should be done more slowly than this experiment. especially the heating process should be done slowly, so the evaporation bed be observed more carefully.