theoretical yield from a reaction with the actual yield. % Yield = Actual yield x100 Theoretical Yield Actual Yield = Mass of product produced through experimentation Theoretical Yield = Mass of product predicted to be produced using stoichiometry A theoretical yield close to 100% is very good. Above 100% means that your product is more massive than it should be. Maybe there is something else there? Below 100% might mean that you are not producing enough product. Some of it is being
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Determining the Stoichiometry of Chemical Reactions Objective In this lab we took Fe(NO3)3 and NaOH and mixed 7 different mole ratios in graduated cylinders to determine what the mole ratio is. We also did the same thing with solutions of CuCl2 and Na3PO4. We determine the mole ratios by graphing the volume of reactant #1 vs. volume of precipitate for each reaction. Data Part 1.) Cylinder 1 2 3 4 5 6 7 Fe(NO3)3‚ 0.1 M‚ ml 5 10 12 15 17 20 24 NaOH‚ 0.1 M‚ ml 55 50 48 45
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! Name: Laura Romero ! 5.04H: Gas Stoichiometry Lab Worksheet Data and Observations: Present all relevant data in a data table below. Include an observations section for any observations that you made during the lab. (5 points) Data Table Mass of magnesium strip (grams) 0.032 g Volume of gas collected (mL) 30mL Barometric pressure (atm) 1.1 atm Room Temperature (°C) 22 °C Vapor pressure of the water (torr) 19.8 torr ! ! Calculations: ! 1. Write the
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CaCl2·2H2O 1 mol CaC03 To double-check‚ we can calculate CaCO3 theoretical yield by using Na2CO3 0.72 g Na2C03 x 1 mol Na2C03 x 1 mol CaC03 x 100 g CaC03 = 106 g Na2C03 1 mol Na2C03 1 mol CaC03 www.LabPaq.com 56 ©Hands-On Labs‚ Inc. Experiment Stoichiometry of a Precipitation Reaction Questions A. From your balanced equation what is the theoretical yield of your product? B. According to your data table‚ what is the actual yield of the product? C. What is the percent yield? D. A perfect percent
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Name: Exp 9: Stoichiometry of a Precipitation Reaction Data Tables: Step 3: Show the calculation of the needed amount of Na2CO3 CaCl2.H2O(aq)= m/M =1/147 =0.0068 mol CaCO3(s)=0.0068*1/1 =0.0068 mol CaCO3 (s)= CaCO3 mol *CaCO3 g =0.0068 mol*100.01 g =0.68 g Convert moles of Na-2CO3 to grams of Na2CO3 = 0.00680 moles Na-2CO3 x 105.99g Na-2CO3 1 mole Na-2CO3 = 0.72g 0.72g of Na-2CO3 to fully react with 1g of CaCl2-.2H2O Step 4: Mass of weighing dish
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Title: Preparation of Cyclohexene from Cyclohexanol Aim: To prepare an alkene‚ cyclohexene‚ by the dehydration of an alcohol‚ cyclohexanol‚ and better understand the processes that take place during this reaction. Introduction: One of the common ways of preparing an alkene is through the dehydration of an alcohol. In this experiment cyclohexanol is dehydrated to prepare cyclohexene while using sulfuric acid as a catalyst. A bromine test can be later done to ensure that the end product is
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Experiment2: Preparation of Dibenzalacetone Aim: Using the cabon-cabon bond making ability in carbonyl chemistry‚ Dibenzalacetone is synthesized from 2 equivalent of benzaldehyde and 1 equivalent of acetone in a base catalyzed reaction. Physical Data1: *detailed risk and safety phrases are attached. substance Hazards‚ risks and safety practices MW (g/mol) Amt. Used Mol. mp (K) bp (K) density(g/cm^3) acetone R11‚ R36‚ R67‚ S9‚ S25‚ S26 58.08 0.24 g 0.004 178.2 329.4 0.79 benzaldehyde R22
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Ms. Azlina Banu‚ Ms. Farhanah‚ Ms.Gurpreet‚ Ms. Jamie‚ Ms. Lau Mei Chien‚ Ms. Lily Lee‚ Ms. Nabilah‚ Mr. Ng Sweet Kin‚ Ms. Phang Ying Ning‚ Ms. Precilla‚ Ms. Rachel Tham‚ Ms. Rajalakshmi‚ Mr. Sivabalan‚ Ms. Tan Lee Siew Tutorial 3: Chapter 3 Stoichiometry and Solution Concentration 1. Balance the following equations: (a) (b) 2. V2O5(s) + CaS(s) CaO(s) + V2S5(s) GaBr3(aq) + Na2SO3(aq) Ga2(SO3)3(aq) + NaBr(aq) 316.0 g of aluminum sulfide‚ Al2S3 reacts with 493.0 g of water‚ H2O. Given the
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and 2-bromo-2-methylpropane. In each of the tubes an even amount of silver nitrate is added. Observations of the reactions precipitate are recorded. Same process is done with 2-chlorobutane‚ 2-iodobutane‚ and 2-bromobutane. Experimental Stoichiometry Compound | Molecularweight | Quantity | Moles | 2-butanol | 74.12 g | 7.4 mL (6.0 g) | 0.081 | Sulfuric acid | 98.08 g | 20 mL (12M) | 0.24 | Ammonium bromide | 97.94 g | 8.0 g | 0.082 | The limiting reactant is 2-butanol. Yield
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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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