Balancing Equations By Tyler Dornbusch and Jason Caws

Directions to Game Click on a number box The higher numbers indicate harder questions Compete with a friend to see who accumulates the most points Questions answered incorrectly will cause the player to lose the number of points indicated in the box How To Balance Equations

Goal: To make all elements in equation equal in numbers of atoms To add numbers of atoms to an element, you must add a coefficient to the element. (Ag = 1 atom of silver, 2Ag = 2 atoms of silver) Also, you must consider the subscript below the element and multiply it by the coefficient that you add. (2H 2 = 4 Hydrogen ions)

How To Balance Equations Continued Example 1: H 2 + O 2  H 2 O There are two atoms of hydrogen and two atoms of oxygen on the reactant (left) side of the equation Now you must add coefficients to balance both sides of the equation 2H 2 + O 2  2H 2 0 A two is added in front of the Hydrogen atom and the water molecule to create two oxygen atoms and four hydrogen atoms on both sides of the equation

Another Example Al + Cl 2  AlCl 3 2Al + 3Cl 2  2AlCl 3 Now we have two aluminum atoms and six chlorine atoms on both sides of the equation

One Last Example Fe + H 3 (PO 4 )  H 2 + Fe(PO 4 ) 2 (When parenthesis are added, multiply the subscript that follows the parenthesis with every element inside of the parenthesis) Therefore, (PO 4 ) 2 has two atoms of Phosphorus and eight ions of Oxygen Fe + 2H 3 (PO 4 )  3H 2 + Fe(PO 4 ) 2 Now there is one atom of Iron, six atoms of Hydrogen, two atoms of Phosphorus, and eight atoms of oxygen on both sides of the equation.

Types of Equations Decomposition: One compound decomposes into two atoms –PCl 5  PCl 3 + Cl 2 Synthesis: Two atoms become an ionic compound –H 2 + O 2  H 2 O

Types of Equations Continued Single Displacement: One element and one compound react to produce a different element and a different compound –H 2 + CuO  Cu + H 2 O Double Displacement: Two compounds produce two different compounds –ZnBr 2 + AgNO 3  Zn(NO 3 ) 2 + AgBr

Types of Equations Continued Combustion: A hydrocarbon reacts with oxygen to produce carbon dioxide and water –CH 4 + O 2  CO 2 + H 2 O

Good Luck Now enjoy playing Jeopardy and test your knowledge on balancing equations. Directions: Click on a number box for a question, and click again to see your answer. –(After each answer, click on the house link to return to the Jeopardy slide for a new choice of question) –If you are lucky, you might get some free points for clicking on the mystery question

Double Displacement Single Displacement Synthesis Decomposition Combustion

100 NaCl + Pb(NO 3 ) 2  PbCl 2 + NaNO 3

100 2NaCl + Pb(NO 3 ) 2  PbCl 2 + 2NaNO 3

200 PbCl 2 + Li 2 SO 4  LiCl + PbSO 4

200 PbCl 2 + Li 2 SO 4  2LiCl + PbSO 4

300 FeCl 3 + NaOH  NaCl + Fe(OH) 3

300 FeCl 3 + 3NaOH  3NaCl + Fe(OH) 3

400 Na 3 (PO 4 ) + CaCl 2  Ca 3 (PO 4 ) 2 + NaCl

400 2Na 3 (PO 4 ) + 3CaCl 2  Ca 3 (PO 4 ) 2 + 6NaCl

500 Ba 3 (PO 4 ) 2 + NaClO 4  Na 3 PO 4 + Ba(ClO 4 ) 2

500 Ba 3 (PO 4 ) 2 + 6NaClO 4  2Na 3 PO 4 + 3Ba(ClO 4 ) 2

100 LiI + Cl 2  LiCl + I 2

100 2LiI + Cl 2  2LiCl + I 2

200 S + O 2  SO 3

200 2S + 3O 2  2SO 3

300 AlBr 3 + Cl 2  AlCl 3 + Br 2

300 2AlBr 3 + 3Cl 2  2AlCl 3 + 3Br 2

400 Zn + HCl  H 2 + ZnCl 2

400 Zn + 2HCl  H 2 + ZnCl 2

500 Cl 2 + KBr  Br 2 + KCl

500 Cl 2 + 2KBr  Br 2 + 2KCl

100 H 2 + Br 2  HBr

100 H 2 + Br 2  2HBr

200 Ag + S  Ag 2 S

200 2Ag + S  Ag 2 S

300 P 4 + O 2  P 2 O 5

300 P 4 + 5O 2  2P 2 O 5

400 Al + Cl 2  AlCl 3

400 2Al + 3Cl 2  2AlCl 3

500 NH 3 + HCl  NH 4 Cl

500 NH 3 + HCl  NH 4 Cl This question is already balanced.

100 CuO  Cu + O 2

100 2CuO  2Cu + O 2

200 H 2 O  H 2 + O 2

200 2H 2 O  2H 2 + O 2

300 Zn(OH) 2  ZnO + H 2 O

300 Zn(OH) 2  ZnO + H 2 O This equation is already balanced.

400 Ni(ClO 3 ) 2  NiCl 2 + O 2

400 Ni(ClO 3 ) 2  NiCl 2 + 3O 2

500 Ag 2 O  Ag +O 2

500 2Ag 2 O  4Ag +O 2

100 CH 4 + O 2  CO 2 + H 2 O

100 CH 4 + 2O 2  CO 2 + 2H 2 O

200 FREE QUESTION: YOU GET A FREE 200 POINTS

200 CONGRATULATIONS!!!!

300 C 4 H 10 + O 2  CO 2 + H 2 O

300 2C 4 H O 2  8CO H 2 O

400 C 6 H 12 O 6 + O 2  CO 2 + H 2 O

400 C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O

500 C 7 H 16 + O 2  CO 2 + H 2 O

500 C 7 H O 2  7CO 2 + 8H 2 O