Chapter 3: Stoichiometry of Formulas and Equations
Chapter 3: Stoichiometry of Formulas and Equations CH 121
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This 0 page Class Notes was uploaded by Amelia Notetaker on Tuesday December 1, 2015. The Class Notes belongs to CH 121 at University of Alabama - Huntsville taught by Pamela D Twigg (P) in Fall 2015. Since its upload, it has received 11 views. For similar materials see GENERAL CHEMISTRY I - 90514 - CH 121 - 02 in Chemistry at University of Alabama - Huntsville.
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Date Created: 12/01/15
Chapter 3 Lecture Notes Chemical equations 0 Concise representation of chemical reaction Reactants on the left 0 Products on the right 0 States are in parentheses Reaction types 0 Combination 2 or more substances form 1 product 0 Decomposition one substance forms two or more 0 Combustion produces C02 and H20 Avogadro39s number 0 6022x10quot23 0 Number of atoms in l mole of Carbon12 12g Percent composition 0 of element number of atoms in elementatomic weightformula weight x 100 Calculating empirical formula from percent composition 0 Convert to grams gt convert to moles gtdivide everything by smallest number of moles gt EF Assume sample has 100 g o If the result from dividing everything by the smallest number of moles is a noninteger divide by a number that will make it a whole number Stoichiometry 0 Given grams of substance A gt molecular mass of A gt moles of A gt coef cient of A and B from equation gt moles of substance B gt molecular mass B gt grams of A Found Limiting reactants 0 Substance completely used up or used up rst is the LR Smallest stoichiometric amount present Theoretical and actual yield 0 Theoretical amount of product that can be made under perfect conditions 0 Actual amount one actually produces and measures
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