Stoichiometry and Limiting Reagents
Stoichiometry and Limiting Reagents CHEM 103 - 002
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CHEM 103 - 002
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This 3 page Class Notes was uploaded by Karlee Nelsen on Friday October 16, 2015. The Class Notes belongs to CHEM 103 - 002 at University of Wisconsin - Eau Claire taught by Sanchita Hati in Fall 2015. Since its upload, it has received 33 views. For similar materials see General Chemistry I in Chemistry at University of Wisconsin - Eau Claire.
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Date Created: 10/16/15
Stoichiometry study of mass relationships in chemical reactions Quantitative information about chemical reactions Uses the concept of The Law of Conservation of Mass Mass before reaction mass after reaction Molar Ratios N2 3H2 9 1 mol N2 1 mol N2 3 mol H2 3 mol H2 2 mol NH3 2 mol NH3 A B 9 C D Finding the mass of D produced from A A x mol A x mol D x molar mass gramsD grams molar mass gramsM mol A mol D example 12 S 3F2 g 9 S How many mols of Iodine are needed to form 4 mols of Iodine Tri uoride Given 4 mol IF3 Find mol 12 1 mol 2 4m0lIF3 XWZ 2m0l12 Limiting Reagent REACTANT that completely reacts with other reactants b until none remains of the limiting reagent remains but there is still left over reactants b Limiting reagent limits the amount of new substance that can be produced EXAMPLE Mgs 2HC1aq 9 MgClz s H2 g 10 mol HCl is placed in each of three asks Mg was also placed in each ask in the amounts of 6g 12g and 18g Which reactants were the limiting reagents and which were in excess Limiting reagent excess reagent 1m0lMg ZmolHCl 1m0lMg ZmolHCl 1m0lMg ZmolHCl And if 243 g Mg react with 120g HCl which is in excess and which is the limiting 1m0lMg ZmolHCl 36461g HCl x x 243gMg 1m0lMg 1m0lHCl 1m0lHCl 1m0lMg 243g Mg X X 36461g HCl ZmolHCl 1m0lMg HCl is in excess because more can react with more Mg than was available in the reaction which 243g Mg x 72922g HCl 120g HCl x 7997g Mg is evident by the smaller number calculated through stoichiometry in relation the amount put into the reaction The starting mass and ending mass in each respective calculation tells the amount of chemicals that will react in proportion with each other so that neither are limiting or in excess ANOTHER EXAMPLE NaHCO3S HClaq 9 NaClaq H200 C02g Calculate mols of reactants Given 5mL 6Molar HCl 2g NaHCO3 2 g NaHC03 x W 024 mol NaHC03 84019 Ncho3 mol 6m0l V L mol 2 M x VL mol 2 x 005L 2 O30m0l HCl 024 IN HCO X 1m0lHCl 024 lHCl 39 m0 a 3 1m0lNaHC03 39 m0 030 lHCl x 1m0lNaHC03 030 IN HCO 39 m0 1m0lHCl 39 m0 a 3 NaHCO3 is the limiting reactant because the stoichiometry produces less HCl than is used in the reaction while HCl calculates to a larger amount of NaHCO3 than is available in the reaction Theoretical Yield is the maximum product yield that can be expected based on the masses of the reactants and the reaction calculated through stoichiometry Actual Yield is the experimentally measured amount of product that results upon completion of the reaction Percent Yield is the measure of the extent of the reaction in terms of actual vs theoretical experimental actual yield 2 100 x theoretical TO FIND THEORETICAL YIELD 1St write equation Au C12 9 AuC13 2nd balance the equation 2Au 3C12 9 2AuC13 10g Au and 10g C12 are combined and begin to react Limiting reactant 3rd do the stoichiometric calculation 1 mol Au 2 mol AuCl3 30335g AuCl3 196967g Au x 2 mol Au x 1 mol AuCl3 1 mol Clz 2 mol AuCl3 30335g AuCl3 709g Clz x 3 mol Clz x 1 mol AuCl3 4th nd the amount in excess 285g 154g 2 131g AuCl3 1 mol AuCl3 3 mol Clz 709g Clz 30335g AuCl3 x 2 mol AuCl3 x 1 mol Clz 10g Au x lOg Clz x 285g AuCl3 131g AuCl3 x 459g Clz in excess
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