1. Balancing Equation Notes

2. Why do we need to be balanced?

3. So all atoms that are on the reactant side are also on the product side – Remember the Conservation of Mass Law

4. Steps to balancing Equations 1.Count the # of atoms of Each Side 2.Add Coefficients where necessary 3.Recount the # of atoms on each side 4.Continue adding coefficients until the number of atoms on each side are equal.

5. Step 1: Count the # of atoms of Each Side Mg = O = Mg = O = Mg + O 2  MgO

6. Step 1: Count the # of atoms of Each Side Mg = 1 O = 2 Mg = 1 O = 1 Mg + O 2  MgO

7. Step 2: Add Coefficients where necessary Are the numbers on both sides even? – NO! Add coefficients Mg = 1 O = 2 Mg = 1 O = 1 Mg + O 2  MgO

8. Step 2: Add Coefficients where necessary Are the numbers on both sides even? – NO! Add coefficients Mg = 1 O = 2 Mg = 1 O = 1 Mg + O 2  2MgO

9. Step 3: Recount the # of atoms on each side Are the atoms even now? – No! Mg = 1 O = 2 Mg = 1 2 O = 1 2 Mg + O 2  2MgO

10. Step 4: Continue adding coefficients until the number of atoms on each side are equal. Mg = 1 O = 2 Mg = 1 2 O = 1 2 Mg + O 2  2MgO

11. Step 4: Continue adding coefficients until the number of atoms on each side are equal. Mg = 1 2 O = 2 Mg = 1 2 O = 1 2 2Mg + O 2  2MgO

12. Are they equal now? – YES – The equation is balanced!!! Step 4: Continue adding coefficients until the number of atoms on each side are equal. Mg = 1 2 O = 2 Mg = 1 2 O = 1 2 2Mg + O 2  2MgO

13. Practice Balancing Equations Na + Cl 2  NaCl

14. Practice Balancing Equations Step 1: Count atoms Na + Cl 2  NaCl Na = 1Na = 1 Cl = 2Cl = 1

15. Practice Balancing Equations Step 2: Add Coefficients Na + Cl 2  2 NaCl Na = 1Na = 1 Cl = 2Cl = 1

16. Practice Balancing Equations Step 3: Recount Atoms Na + Cl 2  2 NaCl Na = 1Na = 1 2 Cl = 2Cl = 1 2

17. Practice Balancing Equations Step 4: Continue adding coefficients until atoms are equal 2 Na + Cl 2  2 NaCl Na = 1 2Na = 1 2 Cl = 2Cl = 1 2

18. Additional Practice CH 4 + O 2  CO 2 + H 2 O Li + HNO 3  LiNO 3 + H 2 Al + O 2  Al 2 O 3

19. Additional Practice CH 4 + O 2  CO 2 + H 2 OC = 1 H = 4H = 2 O = 2O = 3

20. Additional Practice CH 4 + O 2  CO 2 + 2H 2 OC = 1 H = 4H = 2 4 O = 2O = 3 6

21. Additional Practice CH 4 + 3O 2  CO 2 + 2H 2 OC = 1 H = 4H = 2 4 O = 2 6O = 3 6

22. Additional Practice Li + HNO 3  LiNO 3 + H 2Li = 1 H = 1H = 2N = 1O = 3

23. Additional Practice Li + 2HNO 3  LiNO 3 + H 2Li = 1 H = 1 2H = 2 N = 1 2N = 1 O = 3 6O = 3

24. Additional Practice Li + 2HNO 3  2LiNO 3 + H 2 Li = 1Li = 1 2 H = 1 2H = 2 N = 1 2N = 1 2 O = 36O = 3 6

25. Additional Practice 2Li + 2HNO 3  2LiNO 3 + H 2 Li = 12Li = 1 2 H = 1 2H = 2 N = 1 2N = 1 2 O = 36O = 3 6

26. Additional Practice Al + O 2  Al 2 O 3 Al = 1Al = 2 O = 2O = 3

27. Additional Practice 2Al + O 2  Al 2 O 3 Al = 1 2Al = 2 O = 2O = 3

28. Additional Practice 2Al + O 2  2Al 2 O 3 Al = 1 2Al = 2 4 O = 2O = 3 6

29. Additional Practice 2Al + 3O 2  2Al 2 O 3 Al = 1 2Al = 2 4 O = 2 6O = 3 6

30. Additional Practice 4Al + 3O 2  2Al 2 O 3 Al = 1 2 4Al = 2 4 O = 2 6 O = 3 6