Chemical Equations

1
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. 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. 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

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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.

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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.

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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 CaCl2 2 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 CaCl2 2 AgNO3 ? Ca(NO3)2 2 AgCl
• Balance the Ca, Cl, and Ag to complete.

22
Tips for balancing reactions.

1. Start by balancing elements that are only found
   once on each side of the reaction.
2. When the same polyatomic ion is present on both
   sides of an equation, balance the entire ion
   rather than the individual elements.
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.

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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.

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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.

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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.

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Synthesis Reactions (A B ? AB)

• Two or more substances combine to form a larger
  compound.
• Example 2 H2(g) O2(g) ? 2 H2O(l)

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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)

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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.

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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.

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Example 2C3H8(g) O2(g) ? 6CO2(g) H2O(g)

• Start by giving the hydrocarbon a coefficient of
  2.
• Now balance the carbons.

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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.

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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.)

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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
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
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).

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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)