Investigating stoichiometry Essay

In this look into we are going to get a better understanding of chemical core stoichiometry. We are going to be reacting atomic number 11 bicarbonate (NaHCO3) and atomic number 11 carbonate (Na2CO3) with hydrochloric acid (HCl). To start eat up the mass of cardinal unknown region substances (being the sodium bicarbonate and sodium carbonate) allow for be taken. We will require to construct balanced equations for both of the reactants with the HCl and use this to guide us to figure out how much HCl will be shooted to react with each of the unknown substances. When this donation is through we will then need to find the percent reappearance (actual yield/theoretical yield x 100). Determining the mass of sodium chloride at the residuum of the experiment is the actual yield. To find the theoretical yield we will need to heating system the sodium chloride so that all the CO2 evaporates and that will leave us with the theoretical yield of sodium chloride.Chemical ReactionsNaHC O3 + HCl = NaCl + H20 + CO2Na2CO3 + HCl = 2NaCl + H2O + CO2Theoretical yields.15g NaHCO3 (1mol NaHCO3/84.0059g NaHCO3) (1mol HCl/1mol NaHCO3) (58.44gNaCl/1molNaCl) = .1043 g NaCl .15g Na2CO3 (1mol Na2CO3/105.988g Na2CO3) (1mol HCl/1mol Na2CO3) (58.44gNaCl/1molNaCl) = .165 g NaClExperimental Procedure standardisation of Unknown 11) Weigh duplicate 0.15 g samples of unknown 1. Dissolve samples in 100- ml distilled water. 2) minimal brain dysfunction bromocresol green indicator, until the solution turns into blue. Titrate it with HCl until green color is reached. 3) rage and boil out CO2 gently. You should aim a blue color again at the end of this step. Cool to room temperature, and continue titration until color color is reached. Note down the volume seen on the buret. 4) vex the substance again until all the unruffled is g 1 and youre left with salt looking particles inner(a) the beaker. 5) Weigh the beaker with the salt inside of it. Then, bully out the beaker thoroughly an d weigh the emptybeaker again. Use these two fishs to find the mass in grams of the salt like particles that were previously in the beaker.Standardization of Unknown 21) Weigh duplicate 0.15 g samples of unknown 1. Dissolve samples in 100- ml distilled water. 2) Add bromocresol green indicator, until the solution turns into blue. Titrate it with HCl until green color is reached. 3) Heat and boil out CO2 gently. You should obtain a blue color again at the end of this step. Cool to room temperature, and continue titration until yellow color is reached. Note down the volume seen on the buret. 4) Heat the substance again until all the liquid is gone and youre left with salt looking particles inside the beaker. 5) Weigh the beaker with the salt inside of it. Then, clean out the beaker thoroughly and weigh the empty beaker again. Use these two weights to find the mass in grams of the salt like particles that were previously in the beaker. Now to identify which substance is NaHCO3 and whi ch is Na2CO3 pay close attention the mass recorded for the salt like particles that were left in the beaker. The one that is closest to .1043 g is the NaHCO3 and the one closest to .165 g is the Na2CO3.ResultsGrams recorded for Unknown 1- .07 g Unknown 1=NaHCO3 Percent yield= 67% Grams recorded for Unknown 2- .14 g Unknown 2= Na2CO3 Percent yield= 85%DiscussionWhen trying to identify which unknown substance was NaHCO3 and which was Na2CO3. We had to make sure to titrate and process each unknown with the same amount of attention. By over titrating one of the unknowns we could have messed up the experiment and would have had to start over. When heating the liquid on the hot plate, we were making all the CO2 in the liquid evaporate leaving us with just the salt like particles that we were looking for. By deliberateness the salt we were able to justify what the substance was because of previous conversions that we had worked out.ConclusionThe purpose of this experiment was to show how titrating a substance can help to identify that substance. Knowing how to work with stoichiometry equations is also a big part of this lab because we used those equations to help us compare and figure out the identity of the substance. After titrating andevaporating the CO2, the weight of unknown 1 was .07 g which was closest to the theoretical yield of .1043 g. Unknown 1 was identified as NaHCO3. After titrating and evaporating the CO2, the weight of unknown 2 was .14 g which was closest to the theoretical yield of .165 g. Unknown 2 was identified as Na2CO3.