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Objectives -Write a chemical reaction with chemical formulas -Balance chemical reactions following the Law of Conservation of Matter
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Part 7: Balancing Equations
Objectives• -Write a chemical reaction with
chemical formulas• -Balance chemical reactions
following the Law of Conservation of Matter
2
Writing Balanced Equations
To balancing equations:1. Write the formulas for the
reactants and products 2. Balance atoms on each side - using
coefficients in front of formulas***Usually leave H atoms followed by O atoms last!
Formulas• Number before a symbol tell you
how many compound units • 2 H2O = H2O + H2O • Subscripts after a symbol tell how
many atoms of that element are in a compound unit and CAN’T BE CHANGED
• H2O = two H atoms and one O 4
EXAMPLE - the conversion of magnesium to magnesium oxide
Magnesium + Oxygen Magnesium oxide
Balance so each side contains:
2 Mg, 2 O = 2 Mg, 2O
EXAMPLE - the combustion (burning) of methane gas (like in a bunsen burner)
Methane + Oxygen Carbon dioxide + Water
CH4 + O2 CO2 + H2O
Notice that each side contains:1 C, 4 H, 2 O = 1 C, 2 H, 3 O
NOT BALANCED!
EXAMPLE - the combustion (burning) of methane gas (like in a bunsen burner)
Methane + Oxygen Carbon dioxide + Water
CH4 + O2 CO2 + 2H2O
Notice that each side contains:1 C, 4 H, 2 O = 1 C, 4 H, 4 O
NOT BALANCED!
EXAMPLE - the combustion (burning) of methane gas (like in a bunsen burner)
Methane + Oxygen Carbon dioxide + Water
CH4 + 2O2 CO2 + 2H2O
Notice that each side contains:1 C, 4 H, 4 O = 1 C, 4 H, 4 O
BALANCED!
EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen Carbon dioxide + WaterC4H10 + O2 CO2 + H2O
Each side contains: 4 C, 10 H, 2 O = 1 C, 2 H, 3 O
NOT BALANCED!
EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen Carbon dioxide + WaterC4H10 + O2 4CO2 + H2O
Each side contains: 4 C, 10 H, 2 O = 4 C, 2 H, 9 O
NOT BALANCED!
EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen Carbon dioxide + WaterC4H10 + O2 4CO2 + 5H2OEach side contains:
4 C, 10 H, 2 O = 4 C, 10 H, 13 O
NOT BALANCED!
EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen Carbon dioxide + Water2C4H10 + O2 8CO2 + 5H2OEach side contains:
8 C, 20 H, 2 O = 8 C, 10 H, 13 O
NOT BALANCED!
EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen Carbon dioxide + Water2C4H10 + O2 8CO2 + 10H2OEach side contains:
8 C, 20 H, 2 O = 8 C, 20 H, 26 O
NOT BALANCED!
EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen Carbon dioxide + Water2C4H10 + 13O2 8CO2 + 10H2OEach side contains:
8 C, 20 H, 26 O = 8 C, 20 H, 26 O
BALANCED!
EXAMPLE:Barium acetate + Sodium phosphate Barium phosphate + Sodium acetate
Ba(C2H3O2)2 + Na3PO4 Ba3(PO4)2 + NaC2H3O2
You can balance by ions if they are present on both sides of the equation - so each side contains:1 Ba+2, 2 C2H3O2
-1, 3 Na+1, 1 PO4-3 =
3 Ba+2, 1 C2H3O2-1, 1 Na+1, 2
PO4-3
NOT BALANCED!!!
EXAMPLE:Barium acetate + Sodium phosphate Barium phosphate + Sodium acetate
3Ba(C2H3O2)2 + Na3PO4 Ba3(PO4)2 + NaC2H3O2
You can balance by ions if they are present on both sides of the equation - so each side contains:3 Ba+2, 6 C2H3O2
-1, 3 Na+1, 1 PO4-3 =
3 Ba+2, 1 C2H3O2-1, 1 Na+1, 2
PO4-3
EXAMPLE:Barium acetate + Sodium phosphate Barium phosphate + Sodium acetate
3Ba(C2H3O2)2 + Na3PO4 Ba3(PO4)2 + 6NaC2H3O2
You can balance by ions if they are present on both sides of the equation - so each side contains:3 Ba+2, 6 C2H3O2
-1, 3 Na+1, 1 PO4-3 =
3 Ba+2, 6C2H3O2-1, 6 Na+1, 2
PO4-3
EXAMPLE:Barium acetate + Sodium phosphate Barium phosphate + Sodium acetate
3Ba(C2H3O2)2 + 2Na3PO4 Ba3(PO4)2 + 6NaC2H3O2
You can balance by ions if they are present on both sides of the equation - so each side contains:3 Ba+2, 6 C2H3O2
-1, 6 Na+1, 2 PO4-3 =
3 Ba+2, 6C2H3O2-1, 6 Na+1, 2
PO4-3
BALANCED!!
Objectives• -Write a chemical reaction with
chemical formulas• -Balance chemical reactions
following the Law of Conservation of Matter
19
