1. Part 7: Balancing Equations
COVALENT BONDING AND VSEPR
Part 7: Balancing Equations

2. Objectives -Write a chemical reaction with chemical formulas
-Balance chemical reactions following the Law of Conservation of Matter

3. Writing Balanced Equations
To balancing equations:
Write the formulas for the reactants and products
Balance atoms on each side - using coefficients in front of formulas
***Usually leave H atoms followed by O atoms last!

4. 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

5. EXAMPLE - the conversion of magnesium to magnesium oxide
Magnesium Oxygen  Magnesium oxide
Balance so each side contains:
2 Mg, 2 O = Mg, 2O

6. EXAMPLE - the combustion (burning) of methane gas (like in a bunsen burner)
Methane + Oxygen  Carbon dioxide + Water
CH O  CO H2O
Notice that each side contains:
1 C, 4 H, 2 O = 1 C, 2 H, 3 O
NOT BALANCED!

7. EXAMPLE - the combustion (burning) of methane gas (like in a bunsen burner)
Methane + Oxygen  Carbon dioxide + Water
CH O  CO H2O
Notice that each side contains:
1 C, 4 H, 2 O = 1 C, 4 H, 4 O
NOT BALANCED!

8. EXAMPLE - the combustion (burning) of methane gas (like in a bunsen burner)
Methane + Oxygen  Carbon dioxide + Water
CH O  CO H2O
Notice that each side contains:
1 C, 4 H, 4 O = 1 C, 4 H, 4 O
BALANCED!

9. EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen  Carbon dioxide + Water
C4H O2  CO H2O
Each side contains:
4 C, 10 H, 2 O = 1 C, 2 H, 3 O
NOT BALANCED!

10. EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen  Carbon dioxide + Water
C4H O2  4CO H2O
Each side contains:
4 C, 10 H, 2 O = 4 C, 2 H, 9 O
NOT BALANCED!

11. EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen  Carbon dioxide + Water
C4H O2  4CO H2O
Each side contains:
4 C, 10 H, 2 O = 4 C, 10 H, 13 O
NOT BALANCED!

12. EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen  Carbon dioxide + Water
2C4H O2  8CO H2O
Each side contains:
8 C, 20 H, 2 O = 8 C, 10 H, 13 O
NOT BALANCED!

13. EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen  Carbon dioxide + Water
2C4H O2  8CO H2O
Each side contains:
8 C, 20 H, 2 O = 8 C, 20 H, 26 O
NOT BALANCED!

14. EXAMPLE - the combustion of butane (lighter fluid)
Butane + Oxygen  Carbon dioxide + Water
2C4H O2  8CO H2O
Each side contains:
8 C, 20 H, 26 O = 8 C, 20 H, 26 O
BALANCED!

15. Ba(C2H3O2)2 + Na3PO4  Ba3(PO4)2 + NaC2H3O2
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!!!

16. 3Ba(C2H3O2)2 + Na3PO4  Ba3(PO4)2 + NaC2H3O2
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

17. 3Ba(C2H3O2)2 + Na3PO4  Ba3(PO4)2 + 6NaC2H3O2
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

18. 3Ba(C2H3O2)2 + 2Na3PO4  Ba3(PO4)2 + 6NaC2H3O2
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!!

19. Objectives -Write a chemical reaction with chemical formulas
-Balance chemical reactions following the Law of Conservation of Matter