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Chapter 3

by: Biljana Martic
Biljana Martic
SIM-UB (University at Buffalo-The State University of New York)
GPA 3.7
Gen Chem 1
Keister, J B

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About this Document

The notes include example problems of limiting reactant and % yield with instructions on how to do them.
Gen Chem 1
Keister, J B
Class Notes
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This 4 page Class Notes was uploaded by Biljana Martic on Sunday September 27, 2015. The Class Notes belongs to CHE 107LLR at SIM-UB (University at Buffalo-The State University of New York) taught by Keister, J B in Summer 2015. Since its upload, it has received 50 views. For similar materials see Gen Chem 1 in Chemistry at SIM-UB (University at Buffalo-The State University of New York).

Similar to CHE 107LLR at SIM-UB (University at Buffalo-The State University of New York)


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Date Created: 09/27/15
Chapter 3 Chemical Reactions contain reactants and products CH4202 CO22H2O Stoichiometric coefficients are the numbers written in front of the molecule ion and atoms They are important because they help keep the same number of each element on both sides of the reaction There are 4 oxygen molecules on the reactants 02 2 oxygen molecules 2 02 4 oxygen molecules There are 4 oxygen molecules on the products C02 2 oxygen molecules H20 1 oxygen molecule 2 H20 2 oxygen molecules 1 carbon on each side 4 hydrogen on each side Balanced equations are important to uphold the in a chemical reaction mass cannot be created nor destroyed Displacement Reactions a more reactive element displaces a less reactive element from its compound Ex Zn ZS gt 2 ZnS Zinc displaces Hydrogen Double Displacement the cations and anions of two different compounds switch places forming two entirely different compounds Ex AgNO3 NaCl gt AgCI NaN03 Combustion the burning of a substance in the presence of oxygen Ex CH4 2 02 gt C02 2 H20 Combination Reaction two reactants combine to form there are fewer products than reactants EX H2 Ci2 Decomposition Reaction a single compound breaks down into two or more elements or new compounds there are more products than reactants EX 2 H202 2 H2O 02 Average Atomic Mass average mass of isotopes same protons and electrons but diff of neutrons Isotopes of Carbon 12C 98892 13C 1108 Multiply mass by percentage of occurrence 98892x12 01108x1300335 12011 amu lt the averages are the s used on the periodic table Molecular Mass the sum of the atoms in the molecule CH4 carbon amu 1201gmol 1 atom hydrogen amu 101 gmol 4 atoms 1201 gmol 101 gmol x4 1605 gmol Formula Mass the sum of the atomic weights of the atoms in its empirical formula C6H1206 Empirical Formula CH20 carbon amu 1201 gmol 1 atom hydrogen 101 gmol 2 atoms oxygen 1600 gmol 1 atom 1201 gmol 101 gmol x 2 1600 gmol 3003 gmol Avogadro s number NA 6022 x 105 when multiplied by moles gives you the number of molecules Conversion Factors Grams lt Use molar mass gtM0leSlt Use Avogadro s gtMolecules When trying to nd grams of one molecule from the grams of another always convert to moles rst then to grams Ex 650 g of aluminum reacts with an excess of oxygen How many grams of aluminum oxide are formed 4A302 gt2Al203 1st convert Al to moles 2ncl compare moles 3rCI convert back to grams 1 mol 1 atom of Al in equation lmolAl 2m0lAl203 10196g0fA1203 550 0 N X 2698g0fAl X 4m0lAl X lmolA1203 123 g Al203 found on periodic T AI 2698 x I O 1600 x I table grams of T atoms found T on periodic table Limiting Reactant Ex A 613 g sample of MgOH2 reacts with 316 g of HCI according to the reaction MgOH2 I HCI gt MgClz 2 H20 Find the limiting reactant In order to do so take each reactant and convert into moles 1m0lMg0H2 613 g MgOH2 X 583gMg0 12 105 mOI M9OH2 1m0l HCl 315gHCIX 365gHCl Next compare the ratios of moles of each reactant needed Pick one of them to see how much of the other you need then compare that to how much you actually have 2 moles of H Cl needed 105 mol MgOH2 x 1m010fMg0H2 21 mol of HCI needed considering that we only have the HCI is the limiting reactant Since that is your limiting reactant you can use it to see how much Mng can be produced theoretically 1m0l MgCl 2 953 g MgCl 2 X 2m0lHCl X 1m0lMgCl2 413 g MgCIZ Percent Yield Suppose that is a lab a chemist actually obtained 458 g MgClz Yield Actual Yield Theoretical Yield X 100 413gMgCl2 458gMgC12 X 100 Stoichiometry tutorial httpswwwyoutubecomwatchvlVTOSWwEM one of the better ones Limiting ReactantReagent httpswwwyoutubecomwatchvkrioEleegc


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