Intro to Reactions (p. 241 – 250) Ch. 8 – Chemical Reactions Intro to Reactions (p. 241 – 250)

A.Signs of a Chemical Reaction Evolution of heat and light Formation of a gas Formation of a precipitate Color change

B.Law of Conservation of Mass mass is neither created nor destroyed in a chemical reaction total mass stays the same atoms can only rearrange 4 H 2 O 4 H 2 O 36 g 4 g 32 g

C. Chemical Equations A+B  C+D REACTANTS PRODUCTS

C. Chemical Equations p. 246

2H2(g) + O2(g)  2H2O(g) D. Writing Equations Identify the substances involved. Use symbols to show: How many? - coefficient Of what? - chemical formula In what state? - physical state Remember the diatomic elements.

D. Writing Equations 2 Al (s) + 3 CuCl2 (aq)  3 Cu (s) + 2 AlCl3 (aq) Two atoms of aluminum react with three units of aqueous copper(II) chloride to produce three atoms of copper and two units of aqueous aluminum chloride. How many? Of what? In what state? 2 Al (s) + 3 CuCl2 (aq)  3 Cu (s) + 2 AlCl3 (aq)

E. Describing Equations Describing Coefficients: individual atom = “atom” covalent substance = “molecule” ionic substance = “unit” 3CO2  2Mg  4MgO  3 molecules of carbon dioxide 2 atoms of magnesium 4 units of magnesium oxide

E. Describing Equations Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) How many? Of what? In what state? One atom of solid zinc reacts with two molecules of aqueous hydrochloric acid to produce one unit of aqueous zinc chloride and one molecule of hydrogen gas.

II. Balancing Equations (p. 250-254) Ch. 8 – Chemical Reactions II. Balancing Equations (p. 250-254)

Coefficient  subscript = # of atoms A. Balancing Steps 1. Write the unbalanced equation. 2. Count atoms on each side. 3. Add coefficients to make #s equal. Coefficient  subscript = # of atoms 4. Reduce coefficients to lowest possible ratio, if necessary. 5. Double check atom balance!!!

B. Helpful Tips Balance one element at a time. Update ALL atom counts after adding a coefficient. If an element appears more than once per side, balance it last. Balance polyatomic ions as single units. “1 SO4” instead of “1 S” and “4 O”

C. Balancing Example 2 Al + CuCl2  Cu + AlCl3 Al Cu Cl 3 3 2 2  1 1 Aluminum and copper(II) chloride react to form copper and aluminum chloride. 2 Al + CuCl2  Cu + AlCl3 Al Cu Cl 3 3 2 2  1 1 2 3  2  6 3  6   3

III. Types of Chemical Reactions (p. 256 - 267) Ch. 8 – Chemical Reactions III. Types of Chemical Reactions (p. 256 - 267)

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) A. Combustion the burning of any substance in O2 to produce heat A + O2  B CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

A. Combustion 4 2 Na(s)+ O2(g)  Na2O(s) C3H8(g)+ O2(g)  5 3 4 Products: contain oxygen hydrocarbons form CO2 + H2O 4 2 Na(s)+ O2(g)  Na2O(s) C3H8(g)+ O2(g)  5 3 4 CO2(g)+ H2O(g)

B. Synthesis the combination of 2 or more substances to form a compound only one product A + B  AB

B. Synthesis H2(g) + Cl2(g)  2 HCl(g)

Al(s)+ Cl2(g)  2 3 2 AlCl3(s) B. Synthesis Products: ionic - cancel charges covalent - hard to tell Al(s)+ Cl2(g)  2 3 2 AlCl3(s)

AB  A + B C. Decomposition a compound breaks down into 2 or more simpler substances only one reactant AB  A + B

C. Decomposition 2 H2O(l)  2 H2(g) + O2(g)

C. Decomposition 2 2 KBr(l)  K(s) + Br2(l) Products: binary - break into elements others - hard to tell 2 2 KBr(l)  K(s) + Br2(l)

A + BC  B + AC D. Single Replacement one element replaces another in a compound metal replaces metal (+) nonmetal replaces nonmetal (-) A + BC  B + AC

Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s) D. Single Replacement Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)

D. Single Replacement Fe(s)+ CuSO4(aq)  Cu(s)+ FeSO4(aq) Products: metal  metal (+) nonmetal  nonmetal (-) free element must be more active (check activity series) Fe(s)+ CuSO4(aq)  Cu(s)+ FeSO4(aq) Br2(l)+ NaCl(aq)  N.R.

AB + CD  AD + CB E. Double Replacement ions in two compounds “change partners” cation of one compound combines with anion of the other AB + CD  AD + CB

Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq) E. Double Replacement Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)

E. Double Replacement Pb(NO3)2(aq)+ KI(aq)  2 2 PbI2(s)+ KNO3(aq) Products: switch negative ions one product must be insoluble (check solubility table) Pb(NO3)2(aq)+ KI(aq)  2 2 PbI2(s)+ KNO3(aq) NaNO3(aq)+ KI(aq)  N.R.