Stoichiometry is the determination of the proportions in which elements or compounds react with one another. Chemical equations require the mole, mass, and atom ratios to remain constant. A double replacement reaction can occur when two ionic compounds are placed in a solution. The equation used in this experiment is a double replacement reaction because 2 pairs of elements react and swap element pairs (A+X + B+Y→B+X + A+Y). When these ionic compounds react and are filtered, a combination of ions will be insoluble in water, so it filters out of the solution. To produce the 2.00 grams of the compound in this experiment, the chemical equation must be balanced, the precipitate must be identified, each reactant and products molar masses should …show more content…
A digital scale should be used to precisely measure the grams of the needed masses, and 25 mL of water should be poured into 2 beakers, each getting 25 mL. Place a flask under a ring stand containing a funnel with an 11 cm filter paper. Each measured reactant goes into one of the beakers, and then these two beakers’ contents are combined. Pour the mixture of the two reactants slowly and intermittently into the funnel until the mixture is gone and it has filtered all the way through the filter paper and into the …show more content…
The needed mass for reactant A was 3.52 grams and the needed mass for reactant B was 2.12 grams. The moles of product C (CaCO3) were also required to have been calculated using mass- to- mole conversion, but the mass was already established, so its mass was not calculated. Results Calculations used to determine the percent error 2.3 g CaCO3 - 2.00 g CaCO3 2.00 g CaCO3 x 100 = 15 % Reactants A and B’s masses were not similar, but the numbers were also not drastically different. Obviously, product C’s solution is different because moles were calculated as opposed to a mass. Percent error was also calculated, using the masses of the filter paper and the precipitate. To determine the mass of the precipitate, measure the dry filter paper containing the precipitate, and then subtract the mass of the individual filter paper from the mass of the dry filter paper and the precipitate. Once the mass of the precipitate is determined, it is used in the percent error equation, seen above. The predicted mass is subtracted from the mass of the precipitate, divided by the predicted mass, and then multiplied
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.
Limiting reagents can be easily determined on paper with stoichiometry, however, observing how it actually works is essential. This lab is focused on putting stoichiometry to use by determining and observing the limiting reagent in the given, balanced reaction. In the field of chemistry, many want to produce a product by reacting to reactants that will create a certain amount of a product. In order to complete this, a chemist requires a balanced equation that states the exact amounts of reactants required to produce an exact amount of a product.
This is significant because this can reveal important reactions and helps to understand various properties of substances. During this experiment, various chemicals will be mixed and their reactions will be identified along with the changes. The observations of the reactions will be displayed on a chart with three columns labelled; before reaction, after reaction, and proof of chemical reaction for the four reactions. During this experiment,
In this experiment, the student was presented with five of the most common types of chemical reactions and practiced balancing chemical equations. Additionally, the student went through an extensive list of chemical reactions was asked to classify the chemical reactions and balance them. In particular, the student was asked to do this for the reactions of: mixing 3 mL calcium chloride with 2 mL sodium phosphate, adding a few drops of water to a test tube containing 0.5 g copper sulfate, heating 0.5 g of copper(II) hydroxide in a test tube, adding a square piece of zinc to a test tube containing 3 mL 3M hydrochloric acid, mixing 2 mL 3M hydrochloric acid with 2 mL sodium carbonate, adding 0.1 g MnO2 to a test tube with 3 mL hydrogen peroxide, combusting a sodium acetate and ethanol solution, mixing 3 g ammonium chloride and 7 g strontium hydroxide octahydrate together and swirling the solution, mixing 2 mL sulfuric acid and 4 mL sodium hydroxide together,
Lab Report By: Amiya Kamal 7B Purpose: The purpose of this experiment is to analyse different substances and see their reaction with each other substances. This experiment is also to examine the particle theory of each mixture.
Throughout the experiment, copper was altered a total of 5 times, but after the final chemical reaction, solid, elemental copper returned. Each time the solution changed color, a precipitate formed, or when gas appeared, indicated that a chemical reaction was occurring. For the first reaction, copper was added to nitric acid, forming the aqueous copper (II) nitrate (where the copper went), along with liquid water, and
In order to create 5 grams of MgSO4 from MgO (Magnesium Oxide) and H2SO4 (Sulfuric acid), we needed to create a balanced equation to find the amount of other chemicals we would need. The balanced equation was MgO + H2SO4 --> MgSO4 + H2O. After creating a balanced equation, we found the amounts of MgO and H2SO4 using stoichiometry. The amount of Magnesium Oxide was 1.674 grams and the amount of Sulfuric acid was 6.923 milliliters. In order to create as close to 5 grams of MgSO4 as possible, we decided to ignore sig-figs, and go to the hundreds place for the sake of exactness (getting an A).
Introduction: The objective of the experiment is to determine the limiting reagent in a chemical reaction. The principles of stoichiometry and limiting reagents will be used to predict the amount of product formed. The amount of product formed and the change in the color of the solution upon mixing of two reactants are being used to predict the limiting reagent and calculate the theoretical yield in grams. My hypothesis was that with the reaction of the zinc with the copper sulfate solution that it would dissolve the zinc to determine the limiting reagent.
Using the Law of Definite Proportions, the mass of this product was used to determine the number of moles of copper and chlorine in the sample, which led to being able to determine the
Most often the question is a word problem so assume that 10 grams of Al reacts completely with Ba. How many grams of Cl would be produced. Balance the chemical equation, by malign sure you have the same number of atoms on both sides of the reactant and product side of the equation. To simplify you would just use the law of conservation of mass. Theme converts any mass value in the mass problem into moles(use the molar mass to do this).
dynamic equilibrium, exothermic and endothermic For this problem, you need to remember that heat is just a piece of the reaction, either reactant or product depending on whether the overall reaction is exothermic or endothermic so adding heat would decrease K. 55. limiting reagent, stoichiometry Upon determining which one of the given reactants is the limiting reagent, simply use the balanced chemical equation, upon ascertaining that it is in fact balanced to find the desired number of moles of desired product. Would most simple to solve out each number of moles of each given reactant and then take the smallest result because that’s the limiting reagent 56. insoluble and soluble compounds alkali metals are always soluble so NaI splits apart completely in solution so that means that any compound containing Na will be completely soluble so the compound that forms the precipitate has to contain I which has a charge of -1 and the first two have positive charges and compounds containing Pb are generally insoluble 57.
Introduction Stoichiometry is inclined within the many branches of chemistry, as it deals with the relative amount of chemical change between the reactant and the product of compounds and elements. In simpler words, Stoichiometry is used for us to predict how much mass of a substance is required to be used in order to be involved in a chemical reaction. As a student myself, I find Stoichiometry the hardest within all the past third quarter grading lessons as I can’t comprehend complex knowledge at a fast rate, thus I needed time to understand all the concepts regarding Stoichiometry.
This type of lab heavily utilizes the concepts of stoichiometry. Stoichiometry looks into the relationship between the relative amounts of substances that are part of a reaction. You can also use stoichiometry for unit conversions, in this case, moles. Moles are a vital unit of measurement that is used to calculate and compare large quantities of small things. Chemists use moles rather than mass when determining relationships between multiple substances because they work with extremely small particles, so using the mass wouldn’t be productive.
Now, along the long line of the history of chemistry, scientists have used symbols, formulas, and equations to indicate the elements present, the relative amounts of elements, and the variety of combinations of atoms during a chemical change. As a bonus, you can also find out how much product will be formed or how much reactant is needed, based on the masses of the substances involved. The science that deals with the manipulation of said variables is called Stoichiometry. Stoichiometry,
Introduction This module is all about Stoichiometry, composition and reaction. I chose this topic not because I am having a really hard time understanding it, but because I want to have a deeper understanding of the lesson. Before actually doing this, I watched some videos on YouTube in order to look for some techniques I can actually apply in doing some written works and execute as an example for this module. In this module, I will focus more on the different techniques on how to convert unit to unit that I learned from different sources and some I self-learned, and of course some from the discussion done by our teacher.