Stoichiometry Of A Double Displacement Reaction Lab Report

An error that could have been present during the lab includes not letting the zinc react completely with the chloride ions by removing the penny too early from the solution. For instance, the percent error of this lab was 45.6%, which was determined by the subtraction of the theoretical percent of Cu 2.5% and the experimental percent of Cu 3.64% and dividing by the theoretical percent of Cu 2.5%. This experiment showed how reactants react with one another in a solution to drive a chemical reaction and the products that result from the

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.

Calculate the percent yield of alum crystals for each trial. 4. Discuss the sources of error that affected the percent yield. 5. Write the balanced net ionic equations for the following reactions: a. 2 Al(s) + 2 KOH(aq) + 6 H2O(liq) → 2 KAl(OH)4(aq) + 3 H2(g) b. 2KAl(OH)4(aq) + H2SO4(aq) → 2 Al(OH)3(s) + 2 H2O(liq) + K2SO4(aq), Al(OH)4-(aq) + H+(aq) → Al(OH)3(s) + H2O(liq) c. 2Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(liq), Al(OH)3(s) + 3 H+(aq) → Al3+(aq) + 3 H2O(liq) d. Al2(SO4)3(aq) + K2SO4(aq) + 24 H2O(liq) → 2 KAl(SO4)2•12 H2O, K+(aq) + Al3+(aq) + 2 SO42-(aq) + 12 H2O(liq) → KAl(SO4)2•12

In order to find the amount of a product made during a double displacement reaction, the product has to be separated from the solution. From this number of moles of precipitate can be calculated. From there the number of moles of reactants can be calculated using the mole ratios of the particular reaction that occurred. As seen in Table 5 it is shown that by finding out the number of moles of the unknown, the molar mass of the unknown can be calculated. From the found mass of the unknown compound, the mound of the original ion can be found.

The possibilities for the identity of the metal include copper and iron (1). The possible charges for copper are 1+ and 2+ (2). The possible charged of iron are 2+ and 3+ (2). By using stoichiometry, it can be concluded that there are only four possible equations: CuNO3 (aq) + NaOH (aq) > NaNO3 (aq) + CuOH

This lab’s end result was to correctly identify each unknown solution using prior knowledge of chemical properties and the results of the first experiment conducted. Unknown solution D was the only colored solution, being blue while the others were clear. This made it easy to then match D up to Copper Sulfate because of its color. As unknown A and B were added together, lots of gaseous bubbles formed and revealed the fact that that reaction was the reaction between Hydrochloric Acid and Sodium Carbonate because it was the only reaction that produced a gas release. Unknown A and C produced the only yellow, brown precipitate just as the reaction between Sodium Carbonate and Silver Nitrate had previously.

To better understand this law, Cu(s) was transformed with different reactions only to return back to Cu s). The initial and final mass of Cu(s) was recorded to give the percent recovery of copper product at

Covert to moles of copper Sulfate  =2.255 ±0.012 / 159.6 = 0.0141 ±0.012 mol CuSO4 Finally simplify those values by dividing by the smallest number  =0.0141 ±0.012 / 0.0141 ±0.012 = 1 Then: 0.100 ±0.020 /0.0141 ±0.012 = 8 3) Molecular Formula= CuSO4 x 8H2O

Purpose: The purpose of performing the 2 gram lab was to obtain 2.00 grams of our solid product, barium sulfate. In order for barium sulfate to be a product, we decided to perform a double replacement reaction. Background: In order for barium sulfate to be a product, we decided to perform a double replacement reaction.

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%.

Stoichiometry is a method used in chemistry that involves using relationships between reactants and products in a chemical reaction, to determine a desired quantitative data. The purpose of the lab was to devise a method to determine the percent composition of NaHCO3 in an unknown mixture of compounds NaHCO3 and Na2CO. Heating the mixture of these two compounds will cause a decomposition reaction. Solid NaHCO3 chemically decomposes into gaseous carbon dioxide and water, via the following reaction: 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g). The decomposition reaction was performed in a crucible and heated with a Bunsen burner.

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.

(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.

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.

(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