1. combustion of a hydrocarbon
Chemical Equations
synthesis
decomposition
combustion of a hydrocarbon
single replacement
double replacement

2. Parts of a chemical equation

3. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)

4. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)

5. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)

6. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)

7. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)

8. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)
s = solid = insoluble = precipitate

9. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)
s = solid = insoluble = precipitate
aq = aqueous = solution = dissolved in water

10. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)
s = solid = insoluble = precipitate
aq = aqueous = solution = dissolved in water
g = gas
l = liquid

11. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)

12. Parts of a chemical equation
Mg(s) + 2 HCl(aq)  H2(g) + MgCl2(aq)

13. In a chemical reaction, existing atoms are simply rearranged
In a chemical reaction, existing atoms are simply rearranged. The atoms on both sides of the equation must balance.
N2 + H2  NH3 (not balanced)

14. In a chemical reaction, existing atoms are simply rearranged
In a chemical reaction, existing atoms are simply rearranged. The atoms on both sides of the equation must balance.
N2 + 3 H2  2 NH3 (balanced)

15. Tips for balancing reactions.
Start by balancing elements that are only found once on each side of the reaction.
Example: CO2 + H2  C2H6 + H2O

16. Tips for balancing reactions.
Start by balancing elements that are only found once on each side of the reaction.
Example: CO2 + H2  C2H6 + H2O
Balance the carbons and oxygens first.

17. Tips for balancing reactions.
Start by balancing elements that are only found once on each side of the reaction.
Example: 2 CO2 + H2  C2H6 + 4 H2O
Balance the carbons and oxygens first.

18. Tips for balancing reactions.
Start by balancing elements that are only found once on each side of the reaction.
Example: 2 CO2 + 7 H2  C2H6 + 4 H2O
Finally, balance the hydrogens.

19. Tips for balancing reactions.
Start by balancing elements that are only found once on each side of the reaction.
When the same polyatomic ion is present on both sides of an equation, balance the entire ion rather than the individual elements.
Example: CaCl2 + AgNO3  Ca(NO3)2 + AgCl

20. Tips for balancing reactions.
Start by balancing elements that are only found once on each side of the reaction.
When the same polyatomic ion is present on both sides of an equation, balance the entire ion rather than the individual elements.
Example: CaCl AgNO3  Ca(NO3)2 + AgCl
The nitrate ions can be balanced as if they are a single element.

21. Tips for balancing reactions.
Start by balancing elements that are only found once on each side of the reaction.
When the same polyatomic ion is present on both sides of an equation, balance the entire ion rather than the individual elements.
Example: CaCl AgNO3  Ca(NO3)2 + 2 AgCl
Balance the Ca, Cl, and Ag to complete.

22. Tips for balancing reactions.
Start by balancing elements that are only found once on each side of the reaction.
When the same polyatomic ion is present on both sides of an equation, balance the entire ion rather than the individual elements.
If you have an element that is even on one side and odd on the other, place a coefficient of 2 in front of the substance on the odd side.

23. There are 3 oxygen atoms on the left and 2 on the right.
3. If you have an element that is even on one side and odd on the other, place a coefficient of 2 in front of the substance on the odd side.
KClO3 → KCl + O2
There are 3 oxygen atoms on the left and 2 on the right.

24. 2KClO3 → KCl + O2 Place a coefficient of 2 in front of the KClO3.
If you have an element that is even on one side and odd on the other, place a coefficient of 2 in front of the substance on the odd side.
2KClO3 → KCl + O2
Place a coefficient of 2 in front of the KClO3.

25. 2KClO3 → 2KCl + 3O2 Now balance the equation.
If you have an element that is even on one side and odd on the other, place a coefficient of 2 in front of the substance on the odd side.
2KClO3 → 2KCl + 3O2
Now balance the equation.

26. Synthesis Reactions (A + B  AB)
Two or more substances combine to form a larger compound.

27. Synthesis Reactions (A + B  AB)
Two or more substances combine to form a larger compound.
Example: 2 H2(g) + O2(g)  2 H2O(l)

28. Synthesis Reactions (A + B  AB)
Two or more substances combine to form a larger compound.
Example: 2 H2(g) + O2(g)  2 H2O(l)
Example: MgO(s) + CO2(g)  MgCO3(s)

29. Decomposition (AB  A + B)
A single compound breaks down into two or more smaller substances.

30. Decomposition (AB  A + B)
A single compound breaks down into two or more smaller substances.
Example: 2 NaHCO3(s)  CO2(g) + H2O(g) + Na2CO3(s)

31. Combustion of a hydrocarbon (CxHyOz + O2  CO2 + H2O)
Any hydrocarbon that burns in the presence of oxygen will produce carbon dioxide and water vapor.

32. Combustion of a hydrocarbon (CxHyOz + O2  CO2 + H2O)
Any hydrocarbon that burns in the presence of oxygen will produce carbon dioxide and water vapor.
Balance by giving the hydrocarbon a coefficient of 2, then balance C, H, and then O. Check to see if the coefficients can be simplified.

33. Combustion of a hydrocarbon (CxHYOz + O2  CO2 + H2O)
Any hydrocarbon that burns in the presence of oxygen will produce carbon dioxide and water vapor.
Balance by giving the hydrocarbon a coefficient of 2, then balance C, H, and then O. Check to see if the coefficients can be simplified.
Example: C3H8(g) + O2(g)  CO2(g) + H2O(g)

34. Example: 2C3H8(g) + O2(g)  CO2(g) + H2O(g)
Start by giving the hydrocarbon a coefficient of 2.

35. Example: 2C3H8(g) + O2(g)  6CO2(g) + H2O(g)
Start by giving the hydrocarbon a coefficient of 2.
Now balance the carbons.

36. Example: 2C3H8(g) + O2(g)  6CO2(g) + 8H2O(g)
Start by giving the hydrocarbon a coefficient of 2.
Now balance the carbons.
Balance the hydrogens.

37. Example: 2C3H8(g) + 10O2(g)  6CO2(g) + 8H2O(g)
Start by giving the hydrocarbon a coefficient of 2.
Now balance the carbons.
Balance the hydrogens.
Balance the oxygens (be sure to count all of them.)

38. Example: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Start by giving the hydrocarbon a coefficient of 2.
Now balance the carbons.
Balance the hydrogens.
Balance the oxygens (be sure to count all of them.)
Finally, if all the coefficients are even, divide them all by 2.

39. combustion of a hydrocarbon
Chemical Equations
synthesis
decomposition
combustion of a hydrocarbon
single replacement
double replacement

40. Single Replacement (A + BC → B + AC)
A pure element replaces a similar element in a compound.

41. Single Replacement (A + BC → B + AC)
A pure element replaces a similar element in a compound.
Metals replace the positive ion, nonmetals replace the negative ion.

42. Single Replacement (A + BC → B + AC)
A pure element replaces a similar element in a compound.
Metals replace the positive ion, nonmetals replace the negative ion.
Example: Ca(s) + AlCl3(aq) →

43. Single Replacement (A + BC → B + AC)
A pure element replaces a similar element in a compound.
Metals replace the positive ion, nonmetals replace the negative ion.
Example: Ca(s) + AlCl3(aq) → Al(s) + CaCl2(aq)

44. Single Replacement (A + BC → B + AC)
A pure element replaces a similar element in a compound.
Metals replace the positive ion, nonmetals replace the negative ion.
Example: 3 Ca(s) + 2 AlCl3(aq) → 2 Al(s) + 3 CaCl2(aq)

45. Single Replacement (A + BC → B + AC)
A pure element replaces a similar element in a compound.
Metals replace the positive ion, nonmetals replace the negative ion.
Example: 3 Ca(s) + 2 AlCl3(aq) → 2 Al(s) + 3 CaCl2(aq)
Example: F2(g) + SrBr2(aq) →

46. Single Replacement (A + BC → B + AC)
A pure element replaces a similar element in a compound.
Metals replace the positive ion, nonmetals replace the negative ion.
Example: 3 Ca(s) + 2 AlCl3(aq) → 2 Al(s) + 3 CaCl2(aq)
Example: F2(g) + SrBr2(aq) → Br2(l) + SrF2(aq)

47. Single Replacement (A + BC → B + AC)
A pure element replaces a similar element in a compound.
Metals replace the positive ion, nonmetals replace the negative ion.
Example: 3 Ca(s) + 2 AlCl3(aq) → 2 Al(s) + 3 CaCl2(aq)
Example: F2(g) + SrBr2(aq) → Br2(l) + SrF2(aq)
H, O, N, Cl, Br, I, F form diatomic molecules in their pure form and require a subscript of 2

48. combustion of a hydrocarbon
Chemical Equations
synthesis
decomposition
combustion of a hydrocarbon
single replacement
double replacement

49. Double Replacement (AB + CD → AD + CB)
Two elements in compounds switch places. One of the two products will be a solid precipitate (insoluble compound).

50. Double Replacement (AB + CD → AD + CB)
Two elements in compounds switch places. One of the two products will be a solid precipitate (insoluble compound).
It is important to include states of matter (s, l, g, aq) when writing these equations.

51. Double Replacement (AB + CD → AD + CB)
Two elements in compounds switch places. One of the two products will be a solid precipitate (insoluble compound).
It is important to include states of matter (s, l, g, aq) when writing these equations.
One product will be a solid, called a precipitate. It will be the product with the greatest charges and no nitrate ions, NO3-.

52. sodium chloride + silver nitrate → ?

53. sodium chloride + silver nitrate → ?
The reactants in a double replacement reaction will always be aqueous.

54. sodium chloride + silver nitrate → ?
The reactants in a double replacement reaction will always be aqueous.
NaCl(aq) + AgNO3(aq) →

55. sodium chloride + silver nitrate → ?
The reactants in a double replacement reaction will always be aqueous.
Switching the metals will give you the correct partners for the products.
NaCl(aq) + AgNO3(aq) →

56. sodium chloride + silver nitrate → ?
The reactants in a double replacement reaction will always be aqueous.
Switching the metals will give you the correct partners for the products.
NaCl(aq) + AgNO3(aq) → AgCl( ) + NaNO3( )

57. sodium chloride + silver nitrate → ?
The reactants in a double replacement reaction will always be aqueous.
Switching the metals will give you the correct partners for the products.
The AgCl is the solid, since the other product has a nitrate ion.
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

58. lithium carbonate + calcium bromide → ?

59. lithium carbonate + calcium bromide → ?
Reactants must be aqueous.
Li2CO3(aq) + CaBr2(aq) 

60. lithium carbonate + calcium bromide → ?
Reactants must be aqueous.
Switch the metals to obtain new compounds.
Li2CO3(aq) + CaBr2(aq)  CaCO3( ) + LiBr( )

61. lithium carbonate + calcium bromide → ?
Reactants must be aqueous.
Switch the metals to obtain new compounds.
Calcium carbonate is the solid, because it has charges of 2+ and 2-, as opposed to lithium bromide, which has charges of 1+ and 1-.
Li2CO3(aq) + CaBr2(aq)  CaCO3(s) + LiBr(aq)

62. lithium carbonate + calcium bromide → ?
Reactants must be aqueous.
Switch the metals to obtain new compounds.
Calcium carbonate is the solid, because it has charges of 2+ and 2-, as opposed to lithium bromide, which has charges of 1+ and 1-.
Balance the equation.
Li2CO3(aq) + CaBr2(aq)  CaCO3(s) + 2 LiBr(aq)