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# Class Note for CHM 115 at IPFW 5

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This 9 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Indiana University taught by a professor in Fall. Since its upload, it has received 10 views.

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Date Created: 02/06/15
Stoichiometry Section 36 Stoichiometry A BALANCED chemical equation gives the ratio of the numbers of MOLES MOLECULES and MASSES of reactants and products involved in a reaction Revisit the Automobile Airbag 2 NaN3s a 2 Nas 3 Me To completely fill an automobile air bag after a headon collision 200 moles of N2g are required How many moles of NaN3s must be added to the air bag Amounts of Substances in Chemical Equations Section 37 4PS 5029 9 P4O1os Conversion Factors Three Types of Stoichiometry Calculations a How many grams of P4010 are produced from 620 grams of P b How many grams of 02 are needed to completely react with 620 grams of P c How many grams of P or 02 are required to yield 950 grams of P4010 Amounts of Substances in Chemical Equations Cont 4Ps quot39 5029 9 P4O1os MM 3097 gmol MM 3200 gmol MM 2839 gmol a How many grams of P4010 are produced from 620 grams of P b How many grams of 02 are needed to completely react with 620 grams of P c How many grams of P are required to yield 950 grams of P4010 Amounts of Substances in Chemical Equations Cont Volume of pure substance A Density Mass of A Volume of solution of A Number of particles of A Molaiity gt Stoichiometry Roadmap Wrmass Moles of A Avogadro s number Balanced equatio n 4 Molar mas s Molaiity Moles of B 4 number Avogadk Volume of pure substance B Density Mass of B Volume of solution of B Number of particles of B Limiting Reactants Section 38 Limiting Reactant or Limiting Reagent Excess Reactant or Excess Reagent The moles of product are always determined by the starting moles of the limiting reagent Two methods to determine which reactant is limiting a calculate the yield in moles or grams of one product separately for each reactant and see which reactant produces the least amount of product ie the limiting reactant b assume all of one reactant reacts and calculate how much of the other reactants isare required Limiting Reactants Cont 4Ps quot39 5029 9 P4O1os MM 3097 gmol MM 3200 gmol MM 2839 gmol If 600 grams of Ps and 700 grams of 029 are allowed to react to form P4010 which one is the limiting reactant Method a product yieldquot Method b reactant required Limiting Reactants Cont 4 Ps quot39 5 029 9 P4O1os MM 3097 gmol MM 3200 gmol MM 2839 gmol From the previous page we determined that 029 is the limiting reactant Therefore Ps must be the excess reactant After all 700 grams of 029 are consumed how many grams of Ps are left unreacted Before the reaction started there were 600 grams of Ps Theoretical Yields Section 38 Theoretical Yield Experimental Yield Percent Yield actual experimental yield Yield X 100 calculated theoretical yield What is the theoretical yield of P4O1os if you have 700 grams of 029 the limiting reactant What is the percent yield if only 570 grams of P4010 is recovered after the reaction goes to completion Limiting Reactants and Theoretical Yield Example Silicon carbide is an extremely hard material used in production of some machine tools polishing and drilling equipment It is made by heating silicon dioxide with carbon according to the following reaction Si02s 3Cs 9 SiCs 2C0g MMgmol 6008 1201 4010 2801 If 100 kg of Si02s and 100 kg of Cs are mixed and heated a What is the limiting reactant b What is the theoretical yield in kilograms of SiCs c How many kilograms of the excess reactant remain unreacted

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