Percent Yield Of Aluminum Lab Report

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.

Green with a silver nitrate (AgNO3) solution. Assuming that certain amounts of AgNO3 was dissolved in a particular amount of water to produce the solution. Find out how many moles of Na2CO3 were involved in the silver nitrate reaction first. To my knowledge of Mr. Green’s addition of 1.5 grams of Na2CO3 to the silver nitrate solution comes from the statement. Since Na2CO3 has a molar mass of 105.99 g/mol, the amount of Na2CO3 is: moles of Na2CO3 = mass / molar mass moles of Na2CO3 = 1.5 g / 105.99 g/mol moles of Na2CO3 = 0.0141 mol

A thermometer was also placed in at an angle. After ten minutes the thermometer was adjusted until it was completely straight in the middle of the beaker. Then the hot plate was turned on. One group member monitored the hot plate, while a second group member retrieved a stirring stick to stir in the beaker, this was to help the copper sulfate to dissolve. Two different group members observed what was occurring so that notes could be taken.

As seen in table 1, the theoretical yield was .712 g of C_17 H_19 NO_3. The % yield of this experiment was 7.51 % of C_17 H_19 NO_3. . This low yield can be explained from a poor recrystallization technique combined with potential contamination. Throughout the experiment, the mixture changed color from green, orange, to yellowish lime, and eventually clear.

Theory The equation that was used in this experiment is % recovered = 100% X m/m0. M = mass of copper recovered and m0 = mass of original copper sample.

Copper Cycle Lab Report Ameerah Alajmi Abstract: A specific amount of Copper will undergo several chemical reactions and then recovered as a solid copper. A and percent recovery will be calculated and sources of loss or gain will be determined. The percent recovery for this experiment was 20.46%.

The decomposition of NaHCO3 is an example of Prevention within Green Chemistry principles because all solid waste in this experiment is collected and used again. The only gaseous wastes generated by the reaction in the experiment are carbon dioxide and water, which are benign (Lab 3). The decomposition reaction of NaHCO3, generates virtually no waste, therefore less hazardous chemical syntheses. The byproducts of the reaction are gaseous CO2 and H2O which possess little or no toxicity to human health and the environment, because of the amounts released in this experiment. (Lab 3).

Objective The foremost purpose of this laboratory experiment is to comprehend notion of moles and Avogadro’s number; and also to grasp the idea of converting among mass, moles, and number of atoms in a sample (Hammerschlag 1). Importance Pace University Professor Carroll Zahn allocated a novel assignment to his introductory computer class, in which the results of this assignment greatly astounded both the professor and his class. The subsequent description that was stimulated by the unanticipated results of this miraculous event is the inspiration for the following case study. Perplexing his class, Professor Zahn told them to calculate the cost of a single aluminum atom in a roll of aluminum foil that he had recently purchased.

(150.22g/mol)(3.5 x 10^-3 mol of nucleophile) = 0.525 g Actual yield = 0.441 g, Percent Yield = (0.441g/0.525g) x 100% = 84% 10. Percent recovery from recrystallization = (0.172g/0.441g) x 100% = 38% 11.

Aluminium is in approximately 8.1% of the earths crust and therefore is very abundant. The extraction process of aluminium is high as it is difficult to extract and uses a great deal of energy, around 621 billion kilowatt hours in 2010, but its demand is very high and in 2010 there was 41.4 million tonnes of aluminium extracted. 9) Step 1 Physical separation: Chalcopyrite (CuFeS2) is found inside a rock and that rock is crushed into a ball mill. The ore is then mixed with water and other chemicals to make the copper sulphide minerals hydrophobic.

The final product weight for percent yield was only the solid E product, which missed one half of the final product produce. If both products were weight, the percent yield would have been larger that it was. Instead of 22.33%, it could have been 44.66%. To prove that both products were obtained, but only one of the two products was analyze, a TLC plate of the DCM layer, that contains both products, and of the final product, was obtain.

Ideally, every mole of each reagent would be used up, and theoretical yield, we are assuming that every last mole of the reactants would

Aluminum hydroxide is also used in the production of glass and cement, in the textile industry for the production of waterproof

The crude alginate yield was calculated from the following equation: Alginate yield (%) = [dry wt. of obtained alginate / dry wt. of sample] × 100 1.2. Block distribution and M/G ratio

51 g and 87.2875 + 6.78= 94.01 g Between 80.51 g and 94.01 g there are exactly 37/80 x 100= 46.25%. This is under value for one standard deviation which leads to speculation about if there was a large harvest during the year creating large apples. 2 standard deviations from the mean is 87.2875 - (2 x 6.78)= 73. 73g and 87.625 + (2x 6.78)= 100.85 g.