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!"#$%&'()( '*+,*-( ( ( ( !"#"$%&'()$*%#+,'(-(.+/&/*+,%&(01"2+34$5( 6%#+,"(!/$75#38+(92+41( Concepts to Know ! Define a chemical reaction ! Correctly write a chemical reaction ! Balance reactions by inspection ! Calculate molecular mass for any compound or molecule ! Apply mole ratios within molecules and between molecules. ! Solve stoichiometry problems ! Convert between mass and moles ! Convert between % composition and mass ! Identify limiting reagent ! Calculate percent yield ! Identify reduction and oxidation equations and pick out the compound being reduced or oxidized :( 92+41;(!"#"$%&()$*%#+,(-(.+/&/,+%&(01"2+34$5(:#<(=<;( Need to Memorize 6.02 x 1023 is Avogadro s number. ! Mass due to specific component $ percent composition = # & '100% " Total molar mass of compound % ! actual yield $ && ' 100% percentage yield = ## " theoretical yield % Oxidation is the loss of electrons from an atom. Reducing agents are reduced Reduction is the gain of electrons by an atom. Oxidizing agents are reduced. >( 92+41;(!"#"$%&()$*%#+,(-(.+/&/,+%&(01"2+34$5(:#<(=<;( Writing and Balancing Equations aA (physical state) + bB (state) " cC (state) + dD (state) HOW TO Balance a Chemical Equation Step [1] Write the equation with the correct formulas. • The subscripts in a formula can never be changed to balance an equation, because changing a subscript changes the identity of a compound. Step [2] Balance the equation with coefficients one element at a time. Step [3] Check to make sure that the smallest set of whole numbers is used. 92+41;(!"#"$%&()$*%#+,(-(.+/&/,+%&(01"2+34$5(:#<(=<;( ?( Solve Stoichiometry Problems aA + bB mass A x÷ MM mass B a:b + dD mass C x÷ MM moles A cC " mass D x÷ MM moles B b:c x÷ MM moles C c:d moles D a:c a:d • Limiting Reactant: Compare moles A & moles B after applying mole ratio. • The reactant with the least number of moles AFTER mole ratio considered is the limiting reactant. • Use limiting reactant # moles to determine moles of products that form ! actual yield $ && ' 100% percentage yield = ## " theoretical yield % @( Redox Half Reactions Cu2+ gains 2 e! Zn2+ + Cu Zn + Cu2+ Zn loses 2 e– Each of these processes can be written as an individual half reaction: Oxidation half reaction: Zn Zn2+ + 2 e! loss of e! Reduction half reaction: 92+41;(!"#"$%&()$*%#+,(-(.+/&/,+%&(01"2+34$5(:#<(=<;( Cu2+ + 2e! gain of e! Cu 6 A(