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Stoichiometry and the Chemical Equation

Purpose and Method The purpose of the laboratory is use our knowledge of stoichiometry in a hands-on experience. Stoichiometry refers to the calculations of the different quantities of the reactants and products of chemical reactions. In our experiment, we will use Stoichiometry to figure out the chemical equation of the reaction between hydrogen peroxide and bleach. Experimental Procedure For the experiment, a setup was made as follows: For each trial, the same amount of bleach was used: 4.0 mL. A specific amount of hydrogen peroxide was used, with each progressing trial using one more gram than the previous. After creating the setup shown above, the flask was stirred, causing the vial containing hydrogen peroxide to spill into the bleach. The reaction released oxygen gas that filled up in the 100 mL graduated cylinder. After the reaction finished, we observed the volume of the oxygen gas. With that data, we graphed the volume of oxygen gas produced versus the mass of the hydrogen peroxide used. Also, we will figure out the density of bleach by doing four trials of massing .500 mL of bleach. Data Set A Actual Mass of H2O2 Used (g) mL O2 Produced (mL) Run 1 1.327 27.5 Run 2 2.616 51 Run 3 3.496 62.5 Run 4 4.675 63 Run 5 5.555 63 Run 6 6.421 63 Run 7 7.455 63 Note: for all trials, a volume of 4.0 mL of bleach was used. Mass of .500 mL of Bleach (g) Density of Bleach Solutions (g/mL) 0.478 0.956 0.456 0.912 0.468 0.936 0.485 0.97 Calculations and Graphs Average Density of Bleach = (.956 + .912 + .936 + .970) / 4 = .944 g/mL From analysis of the graph, the equivalence point is (4.0 mL Bleach, 3.496 g H2O2) (From the Lab Manual) Bleach is 5.25% (w/w) NaOCl (From the Lab Manual) Hydrogen Peroxide is 3.00% (w/w) H2O2 4 mL Bleach x (.944 g Bleach / mL Bleach) x (5.25 g NaOCl / 100 g Bleach) x (1 mol NaOCl / 74.44 g NaOCl) = 2.66 e-3 mol NaOCl (3.00 g H2O2 / 100. Solution) x (3.496 g Solut...

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