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

 

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 2O (l)

 

PHASES OF REACTANTS & PRODUCTS

NOTE: Indicate physical state of each species by (phase)

 

e.g. (g) = gas (s) = solid (l)  = liquid (aq.) = aqueous solution

BALANCING CHEMICAL EQUATIONS

Simple Approach

1. ASSIGN COEFFICIENT OF 1 TO MOST COMPLEX COMPOUND2. LOOK FOR ELEMENTS THAT APPEAR ONLY ONCE IN EQN., AND TRY COEFFICENTS

C5 H12 (g) + O 2 (g) CO 2 (g) + H 2O (l)

BALANCE C

(c) Balance C  

C5 H12 (g) + O2 (g) 5 CO2 (g)+ H2O(l)  

BALANCE H

(b) Balance H

 

C5H12 (g) + O2 (g) CO 2 (g) + 6 H2O (l)

BALANCE O

(c) Balance O

 (i) ADD UP O ON RHS = 5 x 2 + 6 x 1 = 16

(ii) O APPEARS AS O2 ON LHS NEED

16/2 x O2

1C5H12 (g) + 16/2 O2 (g) 5 CO2 (g) + 6 H2O(l)

FINAL POLISHING

CONVERT TO WHOLE NUMBERS:

  1C5H12 (g)+8O 2 (g)5CO2 (g)+6H2O(l)

CHECK COEFFICIENTS

LHS RHS

(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

BUT… WHAT DOES THIS EQUATION MEAN?

1 C5 H12 (g) + 8 O2 (g) 5 CO2 (g)+ 6 H 2O(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