1. WRITING CHEMICAL EQUATIONS Reactants (starting materials)  Products (ending materials) (g) = gas(l) = liquid(s) = solid (aq) = aqueous  = heat  = yields (dissolved in water) X  = catalyst+ = combines The number of molecules (moles) involved in the reaction are written in the front of the chemical formula.

2. WRITING CHEMICAL EQUATIONS Reactants (starting materials)  Products (ending materials) 2 Na(s) + Cl 2 (g)  2 NaCl(s) should be read as: 2 moles (or molecules) of sodium metal in its solid state combined with one mole (or molecule) of chlorine gas to produce two moles (or molecules) of sodium chloride in the solid state.

3. CHEMICAL EQUATIONS CHEMICAL EQUATIONS represent chemical reactions which, in turn, are driven by changes like: CHEMICAL EQUATIONS represent chemical reactions which, in turn, are driven by changes like: Change Observation Change Observation l formation of a precipitatesolid is formed l formation of water heat is formed l formation of a gasbubbles formed other changes are: 'Electrochemistryelectrons are transferred 'Thermochemistryheat is transferred

4. CHEMICAL EQUATIONS In a chemical reaction, atoms can be neither created nor destroyed therefore the total number of atoms on the reactant side must equal the total number of atoms on the product side of the equation. This is known as BALANCING AN EQUATION. Once the reactants & products have been identified, remember the relative ratio of one atom to another in a formula can not be altered. H 2 O is the formula for water and it is safe to drink but if it is changed to H 2 O 2 (hydrogen peroxide) it is no longer safe for drinking. It is a completely different substance! Equations must be balanced by changing the number in front of the formula and not by changing the chemical formula.

5. BALANCING CHEMICAL EQUATIONS BALANCING CHEMICAL EQUATIONS Mg + O 2  MgO First list all atoms in order of metals, nonmetals, then “H” & “O” last. Leave the species that is split between more than one compound for last.Mg - 1 O - 2 O - 1 Next, start with the top atom; one Mg on the reactant side and one Mg atom on the product side. The Mg atom is balanced. Now do oxygen, two “O” atoms on the reactant side and one on the product side. The product side needs to change so place a “2” in front of MgO. Remember you can not change the formula. Mg + O 2  2 MgO This now makes the list: Mg - 1Mg - 2 O - 2O - 2 If a two is placed in front of the Mg on the reactant side; 2 Mg + O 2  2 MgOMg - 2 O - 2O - 2 Now the equation is balanced.

6. PRACTICE PROBLEMS BALANCING CHEMICAL EQUATIONS Balance the following chemical equations. The answers are found on the next slide. 1. __Na 2 SO 4 (aq) + __BaCl 2 (aq)  __BaSO 4 (s) + __NaCl(aq) 2. __Al(s) + __H 2 SO 4 (aq)  __Al 2 (SO 4 ) 3 (aq) + __H 2 (g) 3. __NO 2 (g) + __H 2 O(l) + __O 2 (g)  __HNO 3 (aq) 4. __Al(s) + __HCl(aq)  __AlCl 3 (aq) + __H 2 (g) 5. __C 3 H 8 (g) + __O 2 (g)  __CO 2 (g) + __H 2 O(g)

7. BALANCING CHEMICAL EQUATIONS Answers 1. __Na 2 SO 4 (aq) + __BaCl 2 (aq)  __BaSO 4 (s) + _2_NaCl(aq) 2. _2_Al(s) + _3_H 2 SO 4 (aq)  __Al 2 (SO 4 ) 3 (aq) + _3_H 2 (g) 3. _4_NO 2 (g) + _2_H 2 O(l) + __O 2 (g)  _4_HNO 3 (aq) 4. _2_Al(s) + _6_HCl(aq)  _2_AlCl 3 (aq) + _3_H 2 (g) 5. __C 3 H 8 (g) + _5_O 2 (g)  _3_CO 2 (g) + _4_H 2 O(g)

8. PRACTICE PROBLEMS Balance the following molecular equations: 1. __Na 2 CO 3 (aq) + __HNO 3 (aq)  __H 2 CO 3 (aq) + __NaNO 3 (aq) 2. __ KBr(aq) + __I 2 (g)  __ KI(aq) + __Br 2 (l) 3. __AlCl 3 (aq) + __K 3 PO 4 (aq)  __AlPO 4 (s) + __KCl(aq)

9. Answers to PRACTICE PROBLEMS 1. Na 2 CO 3 (aq) + 2 HNO 3 (aq)  H 2 CO 3 (aq) + 2 NaNO 3 (aq) 2. 2 KBr(aq) + I 2 (g)  2 KI(aq) + Br 2 (l) 3. AlCl 3 (aq) + __K 3 PO 4 (aq)  __AlPO 4 (s) + 3 KCl(aq)

10. GROUP STUDY PROBLEMS Balance the following molecular equations: 1. __Na 2 SO 4 (aq) + __AlCl 3 (aq)  __Al 2 (SO 4 ) 3 (aq) + __NaCl(aq) 2. __NaBr(aq) + __Cl 2 (g)  __ NaCl(aq) + __Br 2 (l) 3. __SbCl 3 (aq) + __Na 2 S(aq)  __Sb 2 S 3 (s) + __NaCl(aq) 4. __Mg(OH) 2 (aq) + __ H 2 SO 4 (aq)  __ H 2 O(l) + __MgSO 4 (aq) 5. __C 2 H 4 (g) + __O 2 (g)  __CO 2 (g) + __H 2 O(g)