4. Review of Classifying and Balancing Chemical Equations
Lets take a moment… sit back… relax… and review some previously learned concepts…

5. 5 main types of chemical reactions:1.
Formation/Synthesis 2.
Decomposition/De-formation 3.
Single Replacement 4.
Double Replacement 5.
Hydrocarbon Combustion 6.
Other Combustions

6. 1. Formation/Synthesis Reactions
Reaction in which 2 or more pure elements combine to form a new compound.
General Equation:
A + B  AB
Element + Element  Compound
Example:
Na + Cl  NaCl
(Sodium Chloride) SALT
(Sodium)
(Chlorine)

7. 2. Decomposition/De-Formation
Reaction in which a compound breaks down into the base elements that formed it.
General Equation:
AB  A + B
Compound  Element + Element
Example:
H2O(l)  H2(g) + O2(g)
(Water)
(Hydrogen)
(Oxygen)

8. AB +C  AC + B 3. Single Replacement NaCl(s) + Li(s) LiCl(s) + Na(s)
Reaction in which an element and a compound react. The “partners” are traded and the elements that make up the compound and lone element are substituted.
General Equation:
AB +C  AC + B
Compound + Element  Compound + Element
Example:
NaCl(s) + Li(s) LiCl(s) + Na(s)
(Salt)
(Lithium)
(Lithium Chloride)
(Soduim)

9. AB +CD  AD + CB 4. Double Replacement
Reaction in which two compounds react and “switch partners”. The elements that make up each compound are swapped to form two new compounds.
General Equation:
AB +CD  AD + CB
Compound + Compound Compound + Compound
Example:
NaCl(s) + LiBr(s) LiCl(s) + NaBr(s)
(Salt)
(Lithium
Bromide)
(Lithium
Chloride)
(Sodium
Bromide)

10. 5. Hydrocarbon Combustion
Reaction in which a compound containing Hydrogen and Carbon are burned to produce H2O(g) and CO2(g).
General Equation:
CXHX +O2(g)  H2O(g) + CO2(g)
CxHx + Oxygen Water + Carbon Dioxide
Example:
CH4(g) + O2(g) H2O(g) + CO2(g)
(Methane)
(Oxygen)
(Water
Vapour)
(Carbon
Dioxide)

11. X +O2(g)  XO(g) 6. Other Combustions X+ O2(g) XO General Equation:
Reaction in which a compound OR element is burned forming its most common oxide.
General Equation:
X +O2(g)  XO(g)
X + Oxygen XO
Example:
X+ O2(g) XO

12. Steps to Balance a Chemical Equation
Write out the correct chemical formulae for the products and the reactants. Be sure to include all states of matter!
Balance the atoms or ion present in the greatest number. You may do this by finding the lowest common multiple of the two.
Continue to systematically balance the rest of the atoms or ions.
Check the final equation. Make sure every type of atom or ion balances.

13. Ex. 2NaCl(aq) + MgSO4(aq) → Na2SO4(aq) + MgCl2(s)
The coefficients that are placed in front of the compounds and elements represent the mole-to-mole ratio of the reactants and products.
Ex. 2NaCl(aq) + MgSO4(aq) → Na2SO4(aq) + MgCl2(s)
2 moles of sodium chloride reacts with one mole of magnesium sulfate to produce one mole of sodium sulfate and one mole of magnesium chloride
When you are making quantitative calculations, please use the mole-to-mole ratio in your calculations :

14. RXN Limitations RXNS don’t communicate temp. and pressure
RXNS don’t communicate progress or process
RXNS don’t communicate measurable quantities of substances (can’t really weigh or calculate moles, atoms, molecules, ions, etc. without a calculation first)

15. RXN Assumptions RXNS are spontaneous RXNS are fast
RXNS are quantitative (more than 99% complete)
RXNS are stoichiometric (simple, whole number ratio of chemical amounts of reactants and products

16. Net Ionic Equations
Net ionic equations are useful in that they show only those chemical species participating in a chemical reaction
The key to being able to write net ionic equations is the ability to recognize monoatomic and polyatomic ions, and the solubility rules.

17. Net Ionic Equations Pb(NO3)2(aq) + 2HCl(aq)  PbCl2(s) + 2 HNO3(aq)
Step 1:
Pb(NO3)2(aq) + 2HCl(aq)  PbCl2(s) + 2 HNO3(aq)
Step 2:
Dissociate (split up) and soluble ionic compounds AND ionize (split up) all Strong Acids.
Pb2+(aq) + +
2NO3-(aq) +
2 H3O+(aq)
2 Cl-(aq) 
PbCl2(s) +
2 H3O+(aq) +
2NO3-(aq)

18. Net Ionic Equations Step 3: Step 4:
Pb2+(aq) +
2NO3-(aq) +
2 H3O+(aq) +
2 Cl-(aq) 
PbCl2(s) +
2 H3O+(aq) +
2NO3-(aq)
Total Ionic Equation
Step 3:
Cancel out the spectator ions. (Ions that are the same on both sides)
Step 4:
Write down whatever is left over after canceling. Simplify balancing if necessary
Pb2+(aq) +
2 Cl-(aq) 
PbCl2(s)
Net Ionic Equation

19. BaCl2 + Na2SO4  2 NaCl + BaSO4 (aq) (aq) (aq) (s) Step 1:
Complete Balanced Chemical Equation
What type of reaction?
****Have to use solubility table to determine the state (aq) or (s)****
Double Replacement
Step 1:
DONE

20. BaCl2 + Na2SO4  2 NaCl + BaSO4 (aq) (aq) (aq) (s) Step 2:
Complete Balanced Chemical Equation
Ba2+(aq)
2Cl-(aq)
2Na+(aq)
SO42-(aq)
2Na+(aq)
2Cl-(aq)
BaSO4(s)
Step 2:
Dissociate soluble ionic and ionize strong acids.

21. Ba2+(aq) +
2Cl-(aq) +
2Na+(aq) +
SO4+(aq) 
2Na+(aq) +
2Cl-(aq) +
BaSO4(s)
Step 2:
DONE
Step 3:
Cancel out the spectator ions. (Ions that are the same on both sides)
DONE
Step 4:
Write down whatever is left over after canceling. Simplify balancing if necessary.
DONE
Ba2+(aq) +
SO4+(aq) 
BaSO4(s)

23. Gravimetric Stoichiometry:
The process of calculating the masses involved in a chemical reaction using the balancing numbers and (Need/Got).
Steps for Starting stoichiometry:
Balanced Chemical Equation.
Write information…mols….masse under the chemical/element that it relates to.
***Convert information given in question into mols.***
Find the chemical species the question wants to know about and put a N for (need to find) above it.
Put a G for (got information about this) over the species you have information about from the question.

24. Step 1: Fe2O3(s) + CO(g)  Fe(s) + CO2(g) Step 2: Step 3: 1 3 2 3
Write a Balanced Chemical Equation for the reaction.
n = m M 1
Fe2O3(s) + 3
CO(g)  2
Fe(s) + 3
CO2(g)
n = g
55.85 g/mol
m = ?
m = 100.0g
n =1.79 mol
n = ?
n =1.79 mol
Step 2:
Write the information from the question under the correct element/compound
Step 3:
Convert information to mols.

25. Step 4: N G Fe2O3(s) + CO(g)  Fe(s) + CO2(g) Step 5: 1 3 2 3
Put an N over the species you NEED to find out about.
n = m M N G 1
Fe2O3(s) + 3
CO(g)  2
Fe(s) + 3
CO2(g)
n = g
55.85 g/mol
m = ?
m = 100.0g
n =1.79 mol
n = 1.79 mol
Step 5:
Put a G over the compound you know information about.

26. 2 H2(g) + 1 O2(g)  2 H20(l) Step 1: N G Step 2: Step 3: Step 4:
Hydrogen gas and Oxygen gas react to produce water. How much hydrogen gas would be needed to produce g of water?
Step 1:
Write a Balanced Chemical Equation for the reaction. N G
2 H2(g) + 1 O2(g)  2 H20(l)
m = ?
m = g
n = mol
Step 2:
Write the information from the question under the correct element/compound
Step 3:
Convert information to mols.
Step 4:
Put an N over the species you NEED to find out about.
Step 5:
Put a G over the compound you know information about.

27. Gravimetric Stoichiometry:
The process of calculating the masses involved in a chemical reaction using the balancing numbers and (Need/Got).
Steps for FINISHING stoichiometry:
6. Convert mols of got (G) to mols of need (N) by multiple by the conversion factor (N/G).
7. Convert the mols of need (N) into a mass by multiplying by the molar mass (M) of the needed compound.

28. ( ) N G 1 2 nFE x (1/2) = nFe2O3 (1.79mol) X (1/2) = nFe2O3
Step 6:
Convert mols of got (G) to mols of need (N) by multiple by the conversion factor (N/G). N G 1
Fe2O3(s) + 3
CO(g)  2
Fe(s) + 3
CO2(g)
m = ?
m = 100.0g
n =0.895 mol
n =1.79 mol N G 1 2
( )
nFE x (1/2) = nFe2O3
(1.79mol) X (1/2) = nFe2O3
0.895 mol = nFe2O3

29. ( ) 1 2 n = m M m = nM m = (0.895 mol)( ) 2 x 55.85 g/mol MFe2O3(s) =
Step 7:
Convert mols of need (N) into mass of Need (N) by multiplying by N’s Molar Mass.
n = m M N G 1
Fe2O3(s) + 3
CO(g)  2
Fe(s) + 3
CO2(g)
m = g
m = ?
m = 100.0g
m = nM
m = (0.895 mol)( )
n =0.895 mol
n = mol 1 2
g/mol
( )
2 x g/mol
MFe2O3(s) =
3 x g/mol
g/mol
m = g

30. Hydrogen gas and Oxygen gas react to produce water
Hydrogen gas and Oxygen gas react to produce water. How much hydrogen gas would be needed to produce g of water? N G
2 H2(g) + 1 O2(g)  2 H20(g)
m = g
m = ?
m = g
n = mol
Step 6: Convert mols of got (G) to mols of need (N) by multiple by the conversion factor (N/G).
Step 7: Convert the mols of need (N) into a mass by multiplying by the molar mass (M) of the needed compound.

31. Excess and Limiting Reagents
Excess and limiting reagents refer to the reactant that will run out first and stop more product from forming.
Limiting Reagent
Excess Reagent +
24 Crackers
7 Pieces of Cheese =
Which will run out first?

32. Excess and Limiting Reagents
12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid water. Determine the excess and limiting reagent. Calculate the maximum amount of water that could be produced by reacting these two gases together. G N
H2(g) O2(g)  H2O(g) 2 1 2
m = 12.4 g
m = 12.4 g
m = ?
n = m M
n = m M
n = 6.138….mol
n = mol
***From here on it’s like a double stoichiometry to see which can make more H2O(l)***

33. Excess and Limiting Reagents
12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid water. Determine the excess and limiting reagent. Calculate the maximum amount of water that could be produced by reacting these two gases together. G N
H2(g) O2(g)  H2O(g) 2 1 2
m = 12.4 g
m = 12.4 g
m = ?
n = 6.138….mol
n = mol N G 2
( )
H2 n = 6.138…

34. Excess and Limiting Reagents
12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid water. Determine the excess and limiting reagent. Calculate the maximum amount of water that could be produced by reacting these two gases together.
Limiting
Excess N G
H2(g) O2(g)  H2O(g) 2 1 2
m = 12.4 g
m = 12.4 g
m = ?
n = 6.138….mol
n = mol
O2 n = mol
( ) N G 2 1
H2 n = 6.138…

36. Percent Yield X 100% Actual Amount % Yield = Theoretical Amount
In every reaction you can use stoichiometry to calculate the theoretical amount of product that could be made. (Maximum or Total).
When you actually do an experiment, the actual amount that you are able to make is called the actual amount.
Actual Amount
X 100%
% Yield =
Theoretical Amount

37. Stoichiometry ALWAYS involves mols and a balanced chemical equation.
H2(g) O2(g)  H2O(g) 2 1 2
What information can get us n?
What formula’s do we know that have n in them?

38. Called Gravimetric Stoichiometry.2
H2(g) O2(g)  H2O(g) 1 2
ALREADY DID THIS.
Called Gravimetric Stoichiometry.
n = m M
Use this when we turn a mass into mols and do stoichiometry.
STP
SATP
22.4 mol/L
24.8 mol/L
***Because we can calculate mols we can then do stoichiometry just like before.***
n = 1 mol
V = 22.4 L
n = 1 mol
V = 24.8 L

39. 1 5 3 4
C3H8(g) + O2(g)  CO2(g) + H2O(g) 6 4 10
Even Number

40. 1 5 3 4 C3H8(g) + O2(g)  CO2(g) + H2O(g) ( ) N G n = m M n = 275g
m = 275g
V = 698 L
V = ?
22.4 L
1 mol
n = m M
n =
31.2 mol
STP =
( ) N G 5 1
n = 275g
44.11 g/mol
n =
= X
31.2 mol
6.23 mol
X =
698 L

41. 2 2 1 2 (YOU TRY) ( ) N G n = m M Na(s) + H2O(l)  H2(g) + NaOH(s)
m= 37.1 g
V = 20.0 L
24.8 L
1 mol
m= nM
m= (1.61mol)(22.99g/mol)
m= 37.1 g
SATP =
n = m M
= L X
( ) 2 1
n = 1.61…mol
n = 0.806…mol
X =
mol

42. N 1
N2(g) 3  2 +
NH3(g)
H2(g)
V = ?
P = 450 kPa
T = K
T = 80°C

43. 1 3 2 N2(g) +  NH3(g) H2(g) V = ? m = 7500 g m = 7.5 Kg P = 450 kPa
T = K

44. 1 3 2 N2(g) +  NH3(g) H2(g) V = ? m = 7500 g P = 450 kPa n = 3700 mol
T = K
R = 8.314
n = g
2.02 g/mol
( ) N G 2 3
n =
mol
Have to calculate mols to do stoich.

45. V = ? PV = nRT P = 450 kPa T = 333.15 K R = 8.314 n =
mol
V = ( mol)(8.314)(333.15)
(450 kPa)
V = L of NH3
V = 16 KL of NH3

47. C = n V Solution Stoichiometry:
Stoichiometry ALWAYS involves mols and a balanced chemical equation.
Solution Stoichiometry:
Using the principles that govern concentrations to solve for n (mols).
Once you have moles the stoichiometry is the same.
C = n V
***Once you get mols…the rest is the same.***

48. H2SO4(aq) +
KOH(aq) 
H2O(aq)
+ K2SO4(aq)

49. 1
H2SO4(aq) + 2
KOH(aq)  2 1
H2O(aq)
+ K2SO4(aq)

50. C = n V 1 2 1 2 H2SO4(aq) + KOH(aq)  H2O(aq) + K2SO4(aq) ( ) N G
V = L
V = L
C = ?
C = n V
C = mol/L
n =
n = CV
n = (0.150 mol/L)( L)
( ) N G 1 2
n =
n = mol

51. C = n V C = 0.00119 mol 0.010L C = 0.119 mol/L 1 2 1 2 H2SO4(aq) +G 1
H2SO4(aq) + 2
KOH(aq)  2 1
H2O(aq)
+ K2SO4(aq)
V = L
V = L
C = ?
C = mol/L
C = n V
C = mol/L
n =
C = mol
0.010L
C = mol/L