Verna Wang Hannah Palmer CHEM 101-069 Lab 11-19-16 Stoichiometry and Limiting Reagents Lab Report Purpose: We are using the reaction of sodium hydroxide and calcium chloride to illustrate stoichiometry by demonstrating proportions needed to cause a reaction to take place. Background: Just like a recipe would call for a specific amount of one ingredient to a specific amount of another, stoichiometry is the same exact method for calculating moles in a chemical reaction. Sometimes, we may not have enough of or too much of one ingredient , which would be defined as limiting and excess reagent, respectively. Ideally, every mole of each reagent would be used up, and theoretical yield, we are assuming that every last mole of the reactants would …show more content…
(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 …show more content…
If only one reactant is increased, then the chemical reaction will only produce a certain amount of products after the limiting reagent is used up, and in this experiment, the most mass the reaction could produce was 0.4 grams. Although we kept adding calcium chloride, not adding sodium hydroxide in the same proportions will not yield more product, which is the main goal in conducting this lab. We should have seen a plateau at 0.4 grams to show that the limiting reagent inhibited further Ca(OH)2 production, but we made several mistakes in our experiment, which made the data unusable to conclude. Once again, the data is polluted, so these number are not accurate, but it is the data our group has to work with. The theoretical yield should have been more than the actual yield, and the percentages should have been less than 100. The actual data is the result on our experiment vs theoretical, which is based on the calculations above. I have also learned to pay more attention to draining out all of the product completely before continuing to test the experiment, as any small drop of contaminant can veer our results into a different
Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029
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.
AP Chemistry Semester 1 Final Review 2016 Basics of Chemistry: Name the following compounds BO3 H2S NaOH OF8 PCl6 HNO3 HgNO2 Write the formula for each compound Pentaboron triselenide Sulfuric Acid Carbon Monoxide Lithium Chloride How many moles are in 58.6 g of AgNO3 How many grams are in 2.5 moles of Cl2
Tyler White CHEM151LL 32658 04/01/2018 Different Types Chemical Reaction Types and Equations Purpose: The purpose of this lab experiment is to examine different types of chemical reactions such as Decomposition reaction, Synthesis reactions, Combustion reactions, and different Chemical equations. The experiments were conducted online using Late Nite Labs. Materials: Because the experiments were conducted online there wasn’t any physical use of materials, only digital ones, for these labs to be performed. Only the registration for the website was needed to perform these online labs, as well as a desktop computer.
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 last goal was to determine the percent yield of a product formed during a reaction with the unknown compound. Experimental Design The first day of lab consisted of various preliminary tests that helped identify the unknown compound.
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.
To determine the rate of reaction there are many method to be used for example, measuring the mass after the product has been added and measuring the difference in mass on the duration of a digital scale. Another method, which will be used in this experiment is using a gas syringe to measure the volume of the gas which has been produced. The cylinder inside, will be pushed out to show a quantitative presentation of the volume produced by the reaction. Hypothesis
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 yield was 22.33%. In most cases, that means that a lot of possible product was lost. However, in this case, that was not true. When Benzaldehyde reacted with the Wittig reagent, it produced two products: E-Stilbene and Z-Stilbene. The Z product was a liquid, while the E product was a solid.
CONCLUSION When you put an egg in vinegar, we see that the shell dissolves, but do you ever wonder why? An egg is made mostly out of calcium carbonate which reacts with an ingredient in vinegar, acetic acid. Acetic acid is about 4% of the vinegar and what breaks apart the solid calcium carbonate crystals. The bubbles we see, from the egg, is the carbonate that make carbon dioxide and the other calcium ions float free. This is the equation: CaCO3 (s) + 2 HC2H3O2 (aq)
Ali Atwi : Internal assesment – calculating of the concentration of ethanoic acid in vinegar AIM : To calculate the concentration of ethanoic acid CH3COOH in vinegar using stoichiometric equations, ( Yamaha brand ) Introduction : I personally like to add a little bit of vinegar on my food because it makes it taste better, yet I know that vinegar contains acid, and I also know the consequences of highly concentrated acid intake, like severe itching and stomach ache, vomiting. Venigar contains a small percentage of ethanoic acid Ch3COOH. This practical aims to find out the concentration of the of the vinegar against a standard solution of sodium hydroxide soloution of concentration 0.1 mol dm3 through acid-base titration, the label on the bottle says 6%.
Acids are proton donors in chemical reactions which increase the number of hydrogen ions in a solution while bases are proton acceptors in reactions which reduce the number of hydrogen ions in a solution. Therefore, an acidic solution has more hydrogen ions than a basic solution; and basic solution has more hydroxide ions than an acidic solution. Acid substances taste sour. They have a pH lower than 7 and turns blue litmus paper into red. Meanwhile, bases are slippery and taste bitter.
AIM To design an investigation to study the kinetics of a reaction of your choice RESEARCH QUESTION With respect to hydrochloric acid (HCl), what is the order of reaction in the reaction between HCl and calcium carbonate (CaCO3) determined by changing the concentration of HCl and measuring the volume of carbon dioxide gas (CO2) collected in 30 seconds whilst keeping the mass of the powdered CaCO3 constant and the temperature of the reaction system at 25oC? BACKGROUND INFORMATION Calcium carbonate (CaCO3) is a chemical compound that is commonly found in rocks such as chalk, limestone, marble and travertine in all parts of the world. It also used as a form of medicine as a dietary supplement for a person with insufficient calcium intake because calcium is needed by the body for healthy bones, muscles, nervous system, and heart. CaCO3 is also used as an antacid to relieve
Synopsis This experiment is the determination of Calcium Carbonate (CaCO3) content in toothpaste with the use of back titration while demonstrating quantitative transfer of solids and liquids. A accurately weighed quantity of toothpaste was dissolved in excess volumes of HCl. This solution is then titrated with NaOH to find the volume of the excess HCl. The volume of HCl reacted, which is found by substracting the volume of given HCl with the volume of excess HCl reacted, can be further manipulated with mole fractions to find the mass of CaCO3 and thus the CaCO3 content in toothpastes.