2. The Principle Law of Conservation of Mass: mass is neither created nor destroyed during any chemical reactions. No new atom is created nor old atom destroyed during any chemical reactions. Total number for each element involved in any chemical reactions DOES NOT change, which means the total number of any given element on the reactant side has to match that on the product side.

3. The Application The goal – properly balanced chemical equations that have: perfect match of total number of atoms for each element on both reactant and product sides. Simplest set of integers as coefficients. For any given chemical reaction, elements and compounds involved are set and their formulas can not be altered. Hence, any change to the subscripts is NOT allowed. ONLY COEFFICIENTS ARE ALLOWED TO BE CHANGED!!

4. The Process Write skeleton equation. Note: it is of critical importance that correct formulas to be written for all compounds/elements. Make an atom inventory on each side (reactant or product) for all elements involved in the chemical reaction. Select the element that has different number of atoms from one side to another. Find the least common factors for the two numbers.

5. The Process (cont.) Adjust proper coefficients to get the total number of atoms match from both sides. Each time any coefficient is adjusted, re-do the atom inventory on that side of the chemical equation. Repeat the process until total number of atoms for each element perfectly matches on both sides of the chemical equation.

6. Example I Skeleton Equation: H 2 (g) + O 2 (g)  H 2 O(g) Atom Inventory H O 2 2 1 2 Unbalanced. Choose O to start. 2 [2] [4] Still unbalanced. O is fine now. But H is still out. 2 [2] [4] Matched/Balanced BALANCED NOW!

7. Exercises Balance the following Chemical Equations: a) N 2 + O 2  N 2 O b) H 2 O 2  H 2 O + O 2 c) N 2 + H 2  NH 3

8. Exercises a)N 2 + O 2  N 2 O

9. Exercises (Answers) Balance the following Chemical Equations: a) 2N 2 + O 2  2N 2 O b) 2H 2 O 2  2H 2 O + O 2 c) N 2 + 3H 2  2NH 3

10. Example II Skeleton Equation: C 3 H 8 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) Atom Inventory H O C 8 2 3 2 3 1 Unbalanced. Which element to be the first one? Trick #1: Any element in a chemical equation is your dear friend. Balance it last. O should be balanced last here. Balance H first. [2] [7] [3] 4 [8] [10] [3] 3 [8] [10] [3] 5

11. Example III C 2 H 6 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) Atom Inventory H O C 6 2 2 2 3 1 [2] [3] [2] 3 [6] [7] [2] 2 [6] [7] [2] C and H are balanced. But how about O? How could one get 7 O from the diatomic O2 molecule? The Solution: 7 2 2 7 7 2

12. Example III C 2 H 6 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) Atom Inventory H O C 3 2 [12] [14] [4] Wait a minute! How could a coefficient be a fraction not an integer? 7 2 [12] [14] [4] How to resolve this? Remember, above is a balanced chemical equations. Hence the relation still holds after each side is multiplied by the same factor. { } {} 2 2

13. Example III C 2 H 6 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) Atom Inventory H O C 6 2 4 [12] [14] [4] [12] [14] [4] 7 Horary! The chemical equation now is balanced!

14. Exercises Balance the following Chemical Equations: a) FeS 2 + O 2  Fe 2 O 3 + SO 2 b) C 4 H 10 + O 2  H 2 O + CO 2 c) Si 2 H 3 + O 2  SiO 2 + H 2 O d) Fe 2 O 3 + H 2  Fe + H 2 O e) Fe 2 O 3 + CO  Fe + CO 2 f) N 2 + O 2 + H 2 O  HNO 3

15. Exercises (Answers) Balance the following Chemical Equations: a) 4FeS 2 + 11O 2  2Fe 2 O 3 + 8SO 2 b) 2C 4 H 10 + 13O 2  10H 2 O + 8CO 2 c) 2Si 2 H 3 + 11O 2  8SiO 2 + 6H 2 O d) Fe 2 O 3 + 3H 2  2Fe + 3H 2 O e) Fe 2 O 3 + 3CO  2Fe + 3CO 2 f) 2N 2 + 5O 2 + 2H 2 O  4HNO 3

16. Example IV Skeleton Equation: Ba(ClO 3 ) 2 (aq) + Na 2 SO 4 (aq)  BaSO 4 (p) + NaClO 3 (aq) Atom Inventory Wait. Have you seen the patterns here? Trick #2: Keep polyatomic ions that do not change during the reaction as single units. Cl O Ba Na S 2 10 1 2 1 1 7 1 1 1 Unbalanced. How to start? They are a lot of elements! Polyatomic ions do not change during the reaction.

17. Example IV Ba(ClO 3 ) 2 (aq) + Na 2 SO 4 (aq)  BaSO 4 (p) + NaClO 3 (aq) Atom Inventory Now the equation is a lot easier to balance! ClO 3 Ba Na SO 4 2 1 2 1 1 1 1 1 Treat each polyatomic ion as a single unit. 2 [2] [1] [2] [1]

18. Example V Skeleton Equation: Al(OH) 3 (aq) + H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + H 2 O(l) Atom Inventory Trick #3: water molecule (H 2 O) can be treated as H-OH. OH Al H SO 4 3 1 2 1 1 2 1 3 Unbalanced. Polyatomic ions. Al(OH) 3 (aq) + H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + H-OH(l) Wait. Where’s OH ion? 2 [6] [2] [1] 6 [6] [2] [6] [3] 3 [6] [2] [6] [3]

19. Exercises Balance the following Chemical Equations: a) AgNO 3(aq) + NaCl (aq)  AgCl (s) + NaNO 3(aq) b) Fe(NO 3 ) 2(aq) + KOH (aq)  Fe(OH) 2(s) + KNO 3(aq) c) Pb(NO 3 ) 2(aq) + KI (aq)  PbI 2(s) + KNO 3(aq) d) FeCl 3(aq) + (NH 4 ) 2 CO 3(aq)  Fe 2 (CO 3 ) 3(s) + NH 4 Cl (aq) e) H 3 PO 4(aq) + NaOH (aq)  Na 3 PO 4(aq) + H 2 O (l) f) Na 2 CO 3(aq) + H 2 SO 4(aq)  Na 2 SO 4(aq) + CO 2(g) + H 2 O (l)

20. Exercises (Answers) Balance the following Chemical Equations: a) AgNO 3(aq) + NaCl (aq)  AgCl (s) + NaNO 3(aq) b) Fe(NO 3 ) 2(aq) + 2KOH (aq)  Fe(OH) 2(s) + 2KNO 3(aq) c) Pb(NO 3 ) 2(aq) + 2KI (aq)  PbI 2(s) + 2KNO 3(aq) d) 2FeCl 3(aq) + 3(NH 4 ) 2 CO 3(aq)  Fe 2 (CO 3 ) 3(s) + 6NH 4 Cl (aq) e) H 3 PO 4(aq) + 3NaOH (aq)  Na 3 PO 4(aq) + 3H 2 O (l) f) Na 2 CO 3(aq) + H 2 SO 4(aq)  Na 2 SO 4(aq) + CO 2(g) + H 2 O (l)