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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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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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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silver nitrate to react with the salt from perspiration produced 5.8 x 10-2 grams of silver chloride. How many grams of salt was left by the initial latent fingerprint? Brainstorm the steps to solve this problem as you recall the steps to solving stoichiometry calculations. Be prepared to explain your solution including your decisions with significant figures and scientific notation to increase accuracy. Be sure you complete each step of the process: 1. Write the chemical equation. AgNO3 + NaCl
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something + 2Fe+2 Equation 2: 8H+ + 5Fe+2 + MnO4- --> 5Fe+3 + Mn+2 + 4H2O Equation 3: 6H+ + 2MnO4- + 5H2C2O4 --> 2Mn+2 + 10CO2 + 8H2O Conclusion: Therefore the concluded reaction would be: NH3OH+ + 2Fe+3 --> N2O + 2Fe+2 This was obtained by using stoichiometry half reactions the product of that reaction was determined to be N2O. Some systematic errors could be if the wrong molarity was determined for the permanganate because then that would though off the calculations for the Fe+2 and the rest of the
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Mass (g/mol) Amount used (moles) Melting point (Celsius) Color/observations 2‚3-dibromo-3-phenylpropanoic acid 0.285g 307.97 200.4 – 205.1 White powder Limiting reagent: Trans-cinnamic acid (148.16g/mol) 2.5g/148.16g/mol=0.01687mol Stoichiometry ratio: 1:1 ratio Theoretical yield: Since 0.01687 moles of trans-cinnamic acid therefore following the 1:1 ratio one concludes that 2‚3-dibromo-3-phenylpropanoic acid will be of 0.01687 moles Molar mass = 307.97 g/mol Mass=(307.97g/mol)(0.01687mol)=5
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