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: CH 4 §Oxygen: O 2 §Carbon dioxide: CO 2 §Water: H 2 O

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

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

For the reaction of methane: CH 4 + O 2  CO 2 + H 2 O The reaction arrow Methane Oxygen Carbon dioxide Water ReactantsProducts

For the reaction of methane: CH 4 + O 2  CO 2 + H 2 O C H H H H O=O  O=C=O O H H

For the reaction of methane: CH 4 + O 2  CO 2 + H 2 O C H H H H O=O  O=C=O O H H Note, there appear to be more oxygen atoms, fewer hydrogen atoms at the end that at the beginning! 2 H atoms, 3 oxygen atoms 4 H atoms, 2 oxygen atoms

For the reaction of methane: CH 4 + O 2  CO 2 + H 2 O C H H H H O=O  O=C=O O H H There are enough H atoms in one methane molecule to make two water molecules. O H H 2 …but, oxygen is worse off!

For the reaction of methane: CH O 2  CO H 2 O C H H H H O=OO=O  O=C=O O H H Adjust the total number of oxygen atoms with more O 2. Now, oxygen is OK. O H H O=OO=O

For the reaction of methane: CH O 2  CO H 2 O C H H H H O=O  O=C=O O H H All atoms are in the same numbers before and after. O H H O=O

Propane reacts similarly with O 2 C 3 H 8 + O 2  CO 2 + H 2 O None of the atoms appear in the same amounts before and after; the equation is UNBALANCED. Propane

Propane reacts similarly with O 2 C 3 H 8 + O 2  CO 2 + H 2 O There are enough H atoms in one propane molecule to make four water molecules. None of the atoms appear in the same amounts before and after; the equation is UNBALANCED.

Propane reacts similarly with O 2 C 3 H 8 + O 2  CO H 2 O There are enough carbon atoms in propane to make three carbon dioxide molecules. Hydrogens are OK, but carbons are unequal before and after, still.

Propane reacts similarly with O 2 C 3 H 8 + O 2  3 CO H 2 O Adjust the total number of oxygen molecules... Carbons are OK, but oxygen atoms are still unequal before and after.

Propane reacts similarly with O 2 C 3 H O 2  3 CO H 2 O The equation is BALANCED. Now, all atoms are equal in number before and after.

Propane reacts similarly with O 2 C 3 H O 2  3 CO H 2 O The equation is BALANCED. Now, all atoms are equal in number before and after.

A “ picture ” of this reaction: C 3 H O 2  3 CO H 2 O C C C H H H H H H H H

A “ picture ” of this reaction: C 3 H O 2  3 CO H 2 O C C C H H H H H H H H O=O

A “ picture ” of this reaction: C 3 H O 2  3 CO H 2 O C C C H H H H H H H H O=O O=C=O O=O

A “ picture ” of this reaction: C 3 H O 2  3 CO H 2 O C C C H H H H H H H H O=O O=C=O H O H H O H H O H H O 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. §NEVER, EVER change chemical formulas (subscripts) §Use coefficients to balance the number of each element. §Always check to see that elements are in same numbers on both sides.

Practice C 6 H 12 O 6 + O 2  H 2 O + CO 2

Practice C 6 H 12 O 6 + O 2  H 2 O + CO 2 Start with either C or H. Oxygen appears in every formula; save it till last.

Practice C 6 H 12 O 6 + O 2  H 2 O + CO 2 There are enough C atoms in C 6 H 12 O 6 to form six CO 2 molecules. 6

Practice C 6 H 12 O 6 + O 2  H 2 O + 6 CO 2 There are enough H atoms in C 6 H 12 O 6 to form six H 2 O molecules. 6

Practice C 6 H 12 O 6 + O 2  6 H 2 O + 6 CO 2 Six O 2 molecules are required to provide enough total oxygen atoms to balance. Six O atoms Twelve O atoms Twelve O atoms, 6 sets of two 6

Check C 6 H 12 O O 2  6 H 2 O + 6 CO 2 18 total O atoms 18 total oxygen atoms

Check C 6 H 12 O O 2  6 H 2 O + 6 CO 2 12 H atoms12 total H atoms

Check C 6 H 12 O O 2  6 H 2 O + 6 CO 2 six C atoms

Your turn... Al 2 O 3 + H 2  H 2 O + Al

Your turn... 1 Al 2 O H 2  3 H 2 O + 2 Al

Your turn... Al 2 O H 2  3 H 2 O + 2 Al Check: l 2 Al atoms l 3 Oxygen atoms l 6 H atoms l 2 Al atoms l 3 Oxygen atoms l 6 H atoms

Your turn... Ca(OH) 2 + HCl  H 2 O + CaCl 2

Your turn... 1 Ca(OH) HCl  2 H 2 O + 1 CaCl 2

Your turn... Check: l 1 Ca atom l 2 oxygen atoms l 2 chlorine atoms l 4 hydrogen atoms l 1 Ca atom l 2 oxygen atoms l 2 chlorine atoms l 4 hydrogen atoms Ca(OH) HCl  2 H 2 O + CaCl 2

Your turn... H 2 O + Mg 3 N 2  Mg(OH) 2 + NH 3

Your turn... 6 H 2 O + 1 Mg 3 N 2  3 Mg(OH) 2 + 2NH 3

Your turn... 6 H 2 O + Mg 3 N 2  3 Mg(OH) NH 3 Check: l 12 H atoms l six O atoms l 3 Mg atoms l 2 N atoms l 12 H atoms l six O atoms l 3 Mg atoms l 2 N atoms

Your turn... NH 3 + O 2  NO + H 2 O

Your turn... 4 NH O 2  4 NO + 6 H 2 O

Your turn... 4 NH O 2  4 NO + 6 H 2 O Check: l 4 N atoms l 12 H atoms l 10 oxygen atoms l 4 N atoms l 12 H atoms l 10 oxygen atoms