Exam Review Sheet #1 Balancing Equations and Simple Stoichiometry Answers are provided on the second sheet. Please try to do the worksheet without referring to them‚ because you’ll be expected to know this stuff the first day of school! Balance the following equations: 1) ___ N2 + ___ F2 ( ___ NF3 2) ___ C6H10 + ___ O2 ( ___ CO2 + ___ H2O 3) ___ HBr + ___ KHCO3 ( ___ H2O + ___ KBr + ___ CO2 4) ___ GaBr3 + ___ Na2SO3 ( ___ Ga2(SO3)3 + ___ NaBr 5) ___ SnO + ___ NF3 ( ___ SnF2
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Problem Set Introduction to Stoichiometry Name: ______________________________________________________ Course/Yr/Section: ____________ Date:_____________ Direction: Analyze and solve each problem carefully‚ write the solution on the space provided. Box your final answer/s. (10pts each) Determine the molar mass of the following: a. Al(NO3)3 b. ZnSO4 c. Ba(C2H3O2)2 d. NaHCO3 CH3COOH Calculate the percentage oxygen in the following compounds: a. Na2CO3 b. BaSO4 c. BaO
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Amanda Tran Date of lab: 04/25/05 Date submitted: 05/09/05 Chem 2130-3 Experiment 3: Synthesis of Co(acac-NO2)3 Introduction In this lab‚ Co(acac·NO2)3 is synthesized using the Co(acac)3 complex produced in Experiment 2. The Co(acac)3 complex is used as a reagent instead of acacH because acacH cannot be directly converted to 3-nitroacetylacetone. Since Co(acac)3 is not stable in HNO3‚ Cu(NO3)2 and acetic anhydride are used in this reaction to produce the final product
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Chemistry 2202 Review Final Exam Chapter 2: The Mole 1. Isotope – atoms of an element that have the same number of protons but different numbers of neutrons. Ex. The three forms of oxygen are called oxygen-16‚ oxygen-17‚ and oxygen-18. They all have 8 electrons and are written as 16/8 O (8 protons + 8 neutrons)‚ 17/8 O (8 protons + 9 neutrons)‚ and 18/8 O (8 protons + 10 neutrons). 2. 3. Mass Number – the sum of the protons and neutrons in the nucleus of one atom of a particular
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Mass % Determination of empirical formula and molecular formula Combustion analysis Balancing equations Chapter 4 Limiting reactant Mole to mole conversion from reaction stoichiometry Theoretical yield‚ Percent yield‚ Actual yield Solution concentration (molarity) M = n/V ‚ V always in L M1V1=M2V2 ( dilution calculations) Stoichiometry of reactions in solutions M1V1=M2V2 ( dilution calculations) Ionic reactions (formula unit equation‚ complete ionic and net ionic equation). Solubility rule Types of
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balanced equation‚ the mole relationship can be figured out. The amount of product produced from a given amount of reactants based on the balanced chemical equation is referred to as the theoretical yield. The theoretical yield is based on the stoichiometry of the reaction and ideal conditions in which starting material is consumed completely‚ undesired side reactions do not occur‚ and there are no losses in the work-up procedure. On the other hand‚ the
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balance equations‚ the mole relationship can be figured out. The amount of product produced from a given amount of reactants based on the balance chemical equations is referred to as the theoretical yield. The theoretical yield is based on the stoichiometry of the reactions and ideal conditionsin which starting material is consumed completely‚ undesired side reactions do not occur‚ and there are no losses in the work-up procedure. On the other
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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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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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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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