1. CHEMICAL REACTIONS L.H.S. R.H.S. Reactants Products FROM LAW OF CONSERVATION OF MASS: Must have Same Number of Atoms of Each Type on Both Sides of the Equation

2. SIMPLE COMBUSTION REACTION Example:BURN PROPANE IN AIR (Gas Grill) Propane +Oxygen  Carbon Dioxide+ Water C 5 H 12 (g) + O 2 (g)  CO 2 (g) + H 2 O (l)

3. PHASES OF REACTANTS & PRODUCTS NOTE:Indicate physical state of each species by (phase) e.g. (g) = gas (s) = solid (l) = liquid (aq.) = aqueous solution

4. BALANCING CHEMICAL EQUATIONS Simple Approach 1. ASSIGN COEFFICIENT OF 1 TO MOST COMPLEX COMPOUND 2. LOOK FOR ELEMENTS THAT APPEAR ONLY ONCE IN EQN., AND TRY COEFFICENTS C 5 H 12 (g) + O 2 (g)  CO 2 (g) + H 2 O (l)

5. BALANCE C (c) Balance C C 5 H 12 (g) + O 2 (g)  5 CO 2 (g) + H 2 O (l)

6. BALANCE H (b) Balance H C 5 H 12 (g) +O 2 (g)  CO 2 (g) + 6 H 2 O (l)

7. BALANCE O (c) Balance O (i) ADD UP O ON RHS = 5 x 2 + 6 x 1 = 16 (ii) O APPEARS AS O 2 ON LHS  NEED 16/2 x O 2 1C 5 H 12 (g) + 16/2 O 2 (g)  5 CO 2 (g) + 6 H 2 O (l)

8. FINAL POLISHING CONVERT TO WHOLE NUMBERS: 1C 5 H 12 (g) +8O 2 (g)  5CO 2 (g) +6H 2 O (l)

9. CHECK COEFFICIENTS LHSRHS (1 x 5 C) = 5 C 5 C (1 x 12 H) = 12 H (6 x 2 H) = 12 H (8 x 2 O) = 16 O (5 x 2 O + 6 x 1 O)= 16 O

10. BUT… WHAT DOES THIS EQUATION MEAN? 1 C 5 H 12 (g) + 8 O 2 (g)  5 CO 2 (g) + 6 H 2 O (l) 1. = 1 MOLE of Propane Reacts with 8 MOLES of Oxygen to produce 5 MOLES of Carbon Dioxide + 6 MOLES of Water OR 2. = 1 MOLECULE of Propane Reacts with 8 MOLECULES of Oxygen to produce 5 MOLECULES of Carbon Dioxide + 6 MOLECULES of Water