For example‚ when completing this lab‚ you need to make 2 grams of a product. In this case‚ you need to make 2.00 grams of Copper Phosphate and in order find the grams need for each reactant‚ you need to use Stoichiometry. Stoichiometry is the relationship between the relative quantities of substances taking part in a reaction or forming a compound‚ typically a ratio of whole integers. You need to use numerous conversions‚ atomic weights‚ and mole calculations in order
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include determining the amount of Na2CO3 needed to do a full reaction. This was calculated through stoichiometry calculations: Molar mass was first calculated for CaCl2*2H2O Ca = 40.078g Cl2 = 35.453g*2 = 70.906g 2H2 = 1.00794g*4 = 4.03176g 2O = 15.9994g*2 = 31.9988g 40.078g + 70.906g + 4.03176g + 31.9988g = 147.01456g or 147.0 g CaCl2 1g CaCl2 * 2H2O x (1 mol CaCl2 *2H2O/147g CaCl2 *2H2O) = 0.0068 mol of CaCl2*2H2O Molar mass was then calculated for Na2CO3: Na2 = 22.9898g*2 = 45.9796g
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experimental research‚ the percent oxygen of potassium chlorate can be determined using tactics such as stoichiometry‚ a technique used to determine the amount of substances that are in a reaction. Stoichiometry is an efficient way to determine how much of a certain substance is within a certain compound‚ which is used in many practical ways‚ such as pharmaceutical companies using stoichiometry to determine how much of a particular chemical is needed to use within a drug. However‚ within certain gas
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Stoichiometry February 28th‚ 2013 Abstract: The reactions of the Sodium Hydroxide and two acids‚ Hydrochloric Acid and Sulfuric Acid were performed. The heat given off by these two reactions was used to determine the stoichiometric ratio and the limiting reactants in each experiment. Introduction: Coefficients in a balanced equations show how many moles of each reactant is needed to react with each other and how many moles of each product that will be formed. Stoichiometry allows us to
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Ocean County College Department of Chemistry Stoichiometry of a Precipitation Reaction Submitted by Hendy Zelishovsky Date Submitted: 4/26/2012 Date Performed: 4/25/2012 Lab Section: Chem-180-DL1 Course Instructor: Dr. Cynthia Spencer Purpose
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Further‚ I learned more about the molarity of aqueous solutions‚ and how that quantity‚ along with the volume of the solution‚ can be manipulated to find the exact number of moles in a given volume. In this case‚ experimenters were given aqueous solutions of NaOH and CaCl2 in known molarities and then had them react with one another to yield a precipitate of Ca(OH)2. The precipitate was filtered out of the remaining aqueous solution of stoichiometry. In our case‚ all four tests yielded more mass than
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Stoichiometry lab By: James Stewart Purpose: To calculate mole ratios Introduction: There are two types of chemical analysis; qualitative analysis which is the identification of a substance present in a material‚ and qualitative analysis which measures the amount of the substance. In this lab‚ you will perform a quantitative analysis of a two-step reaction. Copper(II) oxide will be synthesized from a known mass of copper(II) sulfate pentahydrate. Using the relationship of the balanced equation
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Lab Report for Experiment #10 Stoichiometry of a Precipitation Reaction Student’s Name ____________________ Date of Experiment ___________ Date Report Submitted _________________ Title: Purpose: Instructor Changes: Weigh out about 1.7 g of CaCl2·2H2O and record your mass to +/- 0.1 g (for example 1.6 g‚ 1.7 g‚ or 1.8 g). We have made this change so that you will have 2 sig figs in subsequent calculations. Have you made any changes to the procedure? Please explain: Data Tables and Observation:
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Title: Stoichiometry Reaction Objectives: 1. To decompose sodium hydrogen carbonate (sodium bicarbonate) by heating. 2. To accurately measure the degree of completion of the reaction by analysing the solid sodium carbonate product. 3. To calculate amount of product with given amount of reactant. 4. To determine amount of heat release in the reaction. Results: Part 1: Thermal Decomposition of NaHCO3 Materials Mass (g) Clean and dry test tube 15.1632 Clean test tube + NaHCO3 17.1647
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The decomposition of sodium carbonate is definitely‚ Na2CO3(s)+CO2(g)+H2O(g). The was the only equation that matched up exactly with my data in terms of percentage. To start off with‚ when I balanced out the equation‚ I got 2 NaHCO3→ 1 Na2CO3(s)+ 1 CO2(g)+ 1 H2O(g). Therefore when I set up my stoichiometry problem I got 3.2 grams NaHCO3 over 1 x 1 mol NaHCO3 over 84.007g NaHCO3 x 1mol Na2CO3 over 2 mol NaHCO3 x 105.987g Na2CO3 x 1 mol Na2CO3. Hence‚ I multiplied 3.2 x 1 x 1 x 105.987 and got 339
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