Chemical Equations

Chemical Equations Describing Chemical Process

Chemical Equations Identify the substances involved in a chemical process Distinguish between the reactants and products in a chemical process Allow easy determination of quantities of substances involved in chemical processes

Verbal description of a common chemical process: “Methane undergoes combustion in oxygen to form carbon dioxide and water.”

The substances: • methane: CH4 • Oxygen: O2 • Carbon dioxide: CO2 • Water: H2O

Reactant(s) Product(s) The form of an equation:

The form of an equation: Reactant(s) Product(s) One or more substances -- the beginning “stuff”

The form of an equation: Reactant(s) Product(s) Reaction arrow; points toward the product(s) One or more substances -- the beginning “stuff”

The form of an equation: Reactant(s) Product(s) One or more substances -- the final “stuff” Reaction arrow; points toward the product(s) One or more substances -- the beginning “stuff”

Methane Carbon dioxide Water Oxygen The reaction arrow Reactants Products For the reaction of methane: CH4 + O2  CO2 + H2O

H H O H C H H H For the reaction of methane: CH4 + O2  CO2 + H2O  O=C=O O=O

For the reaction of methane: CH4 + O2  CO2 + H2O H  O=C=O H O=O O H C H H 4 H atoms, 2 oxygen atoms H 2 H atoms, 3 oxygen atoms Note, there appear to be more oxygen atoms, fewer hydrogen atoms at the end that at the beginning!

H O H For the reaction of methane: CH4 + O2  CO2 + H2O 2 H O=C=O H O=O  O H C H H H There are enough H atoms in one methane molecule to make two water molecules. …but, oxygen is worse off!

For the reaction of methane: CH4 + 2 O2 CO2 + 2 H2O H  O=C=O H H O=O O O H C H O=O H H H Adjust the total number of oxygen atoms with more O2. Now, oxygen is OK.

For the reaction of methane: CH4 + 2 O2  CO2 + 2 H2O H  O=C=O H H O=O O O H C H O=O H H H All atoms are in the same numbers before and after.

Propane reacts similarly with O2 C3H8 + O2  CO2 + H2O Propane None of the atoms appear in the same amounts beforeand after; the equation is UNBALANCED.

Propane reacts similarly with O2 C3H8 + O2  CO2 + H2O None of the atoms appear in the same amounts before and after; the equation is UNBALANCED. There are enough H atoms in one propane molecule to make four water molecules.

Propane reacts similarly with O2 C3H8 + O2  CO2 + 4 H2O Hydrogens are OK, but carbons are unequal before and after, still. There are enough carbon atoms in propane to make three carbon dioxide molecules.

Propane reacts similarly with O2 C3H8 + O2  3 CO2 + 4 H2O Carbons are OK, but oxygen atoms are still unequal before and after. Adjust the total number of oxygen molecules...

Propane reacts similarly with O2 C3H8 + 5 O2 3 CO2 + 4 H2O Now, all atoms are equal in number before and after. The equation is BALANCED.

Propane reacts similarly with O2 C3H8 + 5 O2  3 CO2 + 4 H2O Now, all atoms are equal in number before and after. The equation is BALANCED.

A “picture” of this reaction: C3H8 + 5 O2  3 CO2 + 4 H2O H H H H C C H C H H H

A “picture” of this reaction: C3H8 + 5 O2  3 CO2 + 4 H2O H H H O=O H C C H O=O C O=O H H H O=O O=O

A “picture” of this reaction: C3H8 + 5 O2  3 CO2 + 4 H2O H O=C=O H H O=O H C O=C=O C H O=O C O=C=O O=O H H H O=O O=O

A “picture” of this reaction: C3H8 + 5 O2  3 CO2 + 4 H2O O H H O=C=O H H H O=O H C O=C=O O C H O=O H C H O=C=O O O=O H H H H H O=O O H H O=O

Suggestions to Balance Equations • Work with elements that appear in the fewest formulas first (in one formula on “each side” of the reaction arrow. • Proceed to elements appearing in greater and greater numbers of formulas. • Always check to see that elements are in same numbers on both sides.

Practice C6H12O6 + O2 H2O + CO2

Practice C6H12O6 + O2 H2O + CO2 Start with either C or H. Oxygen appears in every formula; save it till last.

Practice 6 C6H12O6 + O2 H2O + CO2 There are enough C atoms in C6H12O6 to form six CO2 molecules.

Practice 6 C6H12O6 + O2 H2O + 6 CO2 There are enough H atoms in C6H12O6 to form sixH2O molecules.

Twelve O atoms, 6 sets of two Six O atoms Twelve O atoms Six O atoms Practice 6 C6H12O6 + O2 6 H2O + 6 CO2 Six O2 molecules are required to provide enough total oxygen atoms to balance.

Check 18 total oxygen atoms 18 total O atoms C6H12O6 + 6O2 6 H2O + 6 CO2

Check 12 H atoms 12 total H atoms C6H12O6 + 6 O2 6H2O + 6 CO2

Check six C atoms six C atoms C6H12O6 + 6 O2 6 H2O + 6CO2

Your turn... Al2O3 + H2  H2O + Al

Your turn... 1 Al2O3 + 3 H2  3 H2O + 2 Al

Check: 2 Al atoms 3 Oxygen atoms 6 H atoms 2 Al atoms 3 Oxygen atoms 6 H atoms Your turn... Al2O3 + 3 H2  3 H2O + 2 Al

Your turn... Ca(OH)2 + HCl H2O + CaCl2

Your turn... 1 Ca(OH)2 + 2 HCl 2 H2O + 1 CaCl2

Check: 1 Ca atom 2 oxygen atoms 2 chlorine atoms 4 hydrogen atoms 1 Ca atom 2 oxygen atoms 2 chlorine atoms 4 hydrogen atoms Your turn... Ca(OH)2 + 2 HCl 2 H2O + CaCl2

Your turn... H2O + Mg3N2 Mg(OH)2 + NH3

Your turn... 6H2O + 1 Mg3N23 Mg(OH)2 + 2 NH3

Check: 12 H atoms six O atoms 3 Mg atoms 2 N atoms 12 H atoms six O atoms 3 Mg atoms 2 N atoms Your turn... 6H2O + Mg3N23 Mg(OH)2 + 2 NH3

Your turn... NH3 + O2  NO+ H2O

Your turn... 4NH3 + 5 O2 4 NO+ 6 H2O

Your turn... 4NH3 + 5 O2 4 NO+ 6 H2O Check: • 4 N atoms • 12 H atoms • 10 oxygen atoms • 4 N atoms • 12 H atoms • 10 oxygen atoms

Chemical Equations