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

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

3. 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

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

5. C. Chemical Equations
p. 246

6. 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.

7. 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)

8. 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

9. 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.

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

11. 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!!!

12. 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”

13. 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 
 2
 6
3 
6 
 3

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

15. 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)

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

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

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

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

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

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

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

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

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

25. 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.

26. 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

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

28. 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) 
PbI2(s)+ KNO3(aq)
NaNO3(aq)+ KI(aq) 
N.R.