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Stoichiometry Lab Report

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In order to begin this experiment, first one must find the balanced chemical equation for the reaction which occurs between the aluminum and copper (II) chloride. This balanced equation being 2Al(s)+3CuCl2 (aq)3Cu(s)+2AlCl3 (aq). After finding this equation, one must use the process of stoichiometry in order to find how many grams of aluminum are needed in order to produce 0.15 grams of copper. In this experiment, the purpose was to produce between 0.1 and 0.2 grams of copper, so one should attempt to produce 0.15 grams of copper seeing as it is the average of those two numbers. The first step in the stoichiometric process which one has to complete is finding how many grams of copper are in one mole of copper. This calculation being15 grams …show more content…

After placing the aluminum in said beaker, one should gather a stir stick and stir the solution in order to speed up the reaction. One should stir the solution until there are no visible silver pieces of aluminum left in the beaker of what was the copper (II) chloride solution. When there are no visible silver pieces of aluminum left in the beaker, this means that the reaction has occurred. After the reaction has occurred, one needs to separate the products, which are copper and aluminum chloride, from each other in order to see how much copper was able to be produced. Seeing as the products are a solid in the form of copper and a liquid in the form of aluminum chloride, the process of filtration can be used to separate them. The process of filtration can be used seeing as the copper particles will be caught by the filter because they are too large to pass through it, while the aluminum chloride particles are small enough to pass through the filter. In order to filter the products, one needs to obtain a funnel, filter paper, and a clean …show more content…

After seemingly all of the aluminum chloride has been funneled from the copper, one should take the filter paper out of the funnel and let it dry. One needs to let the filter paper dry before weighing it to see how much copper was produced in order to allow the water to evaporate. If one were to weigh the funnel paper while it was wet, they would not get an accurate measurement of how much copper was produced because the scale would show a reading of the amount of copper as well as the amount of water on the filter paper. Once the filter paper has fully dried, one can then weigh the paper in order to determine how much copper was produced. In order to do this one should take a scale and obtain a clean piece of filter paper along with the one with the copper on it. One should weigh the clean piece of filter paper, then the paper used for the experiment. After obtaining both of these weights, one should subtract the weight of the clean paper from the weight of the paper used in the experiment in order to determine the difference between the two sheets of filter paper. This difference represents the amount of copper which was produced in the

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To calculate the percentage of Cu, we divided the final mass of the penny 0.09 and the initial mass of 2.47 and multiplied by 100. To calculate the percentage of Zn, we divided the final mass of the penny 2.38 and the initial mass of 2.47 and multiplied by 100. During the experiment the hydrochloric acid donated its hydrogen ions in the reaction and then the chloride ions reacted with the zinc ions in the solution. Thus, the zinc dissolved in the highly acidic solution which was caused by the high concentration of H2 ions. Hydrogen gas was generated during the reaction which was seen when bubbles were formed as the penny was dissolved into the beaker.

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Qualitative Data: Prior to combining the solution CuCl2 was a transparent blue and Na2CO3 was clear While combining the solutions CuCl2 and Na2CO3, the solution became a cloudy blue After filtering the a dry blue powder was left inside the filter paper and a clear solution had drained through the paper filter. Calculations: Molar Mass and Number of Moles of Reactants CuCl2 and Na2CO3 Molar Mass of CuCl2 = 63.55 g Cu + (2 35.45g Cl) Molar Mass of CuCl2 = 134.45g CuCl2 Molar Mass of Na2CO3 = (2 22.99g Na) + (3 16.00g O) + 12.01g C Molar Mass of Na2CO3 = 105.99g Na2CO3 Moles of CuCl2 = Mass of CuC =

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After I rinsed the copper I transferred it to a watch glass that had been baked for 20 minutes so it would not contain any liquid so as to not counteract the experiment. After I put the copper on this watch glass I baked the copper in the oven for 20 minutes so there would be no liquid left in that to mess up the mass. After 20 minutes I removed the watch glass with the copper in it and weighed itm making sure to weigh the watch glass

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Students first prepped for the lab by cleaning out the crucible. Three boiling chips were added in the crucible once it was wiped out with a paper towel. The crucible was then placed on a clay triangle two finger widths above the Fischer burner. After 10 minutes of the crucible being directly under the flame, the it was clean and students allowed time for it to cool down. Next, the students from then on used tongs to transport the crucible from weighing it and back to the clay triangle.

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The actual yield of the reaction was 4.411 grams of copper and was obtained through the experiment

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The percent recovery of the copper was calculated using the equation, percent recovery = (the mass of the copper recovered after all the chemical reactions/the initial mass of the copper) x 100. The amount of copper that was recovered was 0.32 grams and the initial mass of the copper was 0.46 grams. Using the equation, (0.32 grams/0.46 grams) x 100 equaled 69.56%. The amount of copper recovered was slightly over two-thirds of the initial amount.

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(0.01 moles of NaOH) x (1 mole Ca(OH)2/ 2 moles of NaOH) = 0.005 moles of Ca(OH)2 Tube 1: (0.0020 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.002 moles of Ca(OH)2 (0.002 moles of Ca(OH)2) x (74.08 grams/mole) = 0.1 grams = theoretical yield Tube 2: (0.0035 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.004 moles of Ca(OH)2 (0.004 moles of Ca(OH)2) x (74.08 grams/mole) = 0.3 grams= theoretical yield Tube 3 (0.0050 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.005 moles of Ca(OH)2 (0.005 moles of Ca(OH)2) x (74.08 grams/mole) = 0.4 grams =theoretical yield Tube

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