1. Introduction to Chemical Principles
Antoine Laurent Lavoisier
Chapter 10: Chemical Calculations Involving Chemical Equations

2. The Law of Conservation of Mass
Mass is neither created nor destroyed in any ordinary chemical reaction.
Therefore, the total mass of reactants is always equal to the total mass of products.

3. Chemical equation
Reactants: are starting substance that undergo change in a chemical reaction. As a chemical reaction proceeds, reactants are consumed and new materials with new chemical properties are produced.
Products: are the substances produced as a result of a chemical reaction.

4. Writing and Balancing Equations

5. In writing chemical equations, only molecular formulas are used, not empirical formulas.
2 C3H7OH + 9 O2
6 CO2 + 8 H2O
Products are written on the right side of the equation. These are created by the chemical change.
Reactants are written on the left side of the equation. These undergo chemical change.
Reactants and products are separated by an arrow pointing to the products.
Different reactants and products are separated from each other using plus signs.

6. 2 C3H7OH + 9 O2
6 CO2 + 8 H2O
Coefficients show the number of molecules or moles of each reactant and product involved in the reaction.
This equation reads: “2 molecules of C3H7OH react with 9 molecules of O2 to produce 6 molecules of CO2 and 8 molecules of water.”

7. In order for a chemical equation to be valid, it must:
2 C3H7OH + 9 O2
6 CO2 + 8 H2O
In order for a chemical equation to be valid, it must:
1) Match the facts: for example, O2, not O or O3 is reacting with C3H7OH.
2) Obey the Law of Conservation of Mass - thus, equations must be balanced.

8. 2 Al + Fe2O3 2 Fe + Al2O3 C11H24 C9H20 + CH2=CH2 2 H2O 2 H2 + O2
Info about reaction conditions may be placed above or below the arrow.
heat is added 
2 Al + Fe2O Fe + Al2O3
heat
C11H C9H20 + CH2=CH2
catalyst
electric
2 H2O H2 + O2
current
light hv
CH4 + Cl CH3Cl + HCl

9. The physical state of a substance may be indicated by the use of symbols in parentheses after the substance. E.g.,
(s) for solid
(l) for liquid
(g) for gas
(aq) for aqueous solution 
2Al Fe2O  2Fe Al2O3
(s)
(s)
(l)
(s)

10. To balance an equation, adjust the number of atoms of each element so that they are the same on each side of the equation.
Equations are balanced only by altering the equation coefficient for each item.
Never, never change a substance’s formula to balance an equation.

11. Steps for Balancing Equations

12. Example: Write a balanced equation showing the decomposition of mercury (II) oxide to yield mercury and oxygen.
Step 1 Write the unbalanced (skeleton) equation.
HgO  Hg + O2
The formulas of the reactants and products must be correct!
The reactants are written to the left of the arrow and the products to the right of the arrow.

13. Element Reactant Side Product Side
Step 2 Identify which elements need to be balanced.
Count and compare the number of atoms of each element on both sides of the equation.
HgO → Hg + O2
Element Reactant Side Product Side Hg
There is one mercury atom on the reactant side and one mercury atom on the product side.
Mercury is balanced.

14. Element Reactant Side Product Side
HgO  Hg + O2
Element Reactant Side Product Side O
There are two oxygen atoms on the product side and there is one oxygen atom on the reactant side.
Oxygen needs to be balanced.

15. Step 3 Balance the equation.
Balance each element one at a time, by placing whole numbers (coefficients) in front of the formulas containing the unbalanced element.
A coefficient placed before a formula multiplies every atom in the formula by that number.

16. Element Reactant Side Product Side
HgO  Hg + O2 2
Element Reactant Side Product Side O
Place a 2 in front of HgO to balance O.
Now there are two oxygen atoms on the reactant side and there are two oxygen atoms on the product side.
Oxygen (O) is balanced.

17. Step 4 Check all other elements after each individual element is balanced to see whether, in balancing one element, another element became unbalanced.

18. Element Reactant Side Product Side
2 HgO  Hg + O2
Element Reactant Side Product Side Hg
There are two mercury atoms on the reactant side and there is one mercury atom on the product side.
Therefore, mercury (Hg) is no longer balanced.

19. Element Reactant Side Product Side
2 HgO  Hg + O2 2
Element Reactant Side Product Side Hg
Place a 2 in front of Hg to balance mercury.
Now there are two mercury atoms on the reactant side and there are two mercury atoms on the product side.
Mercury (Hg) is balanced.

20. Element Reactant Side Product Side
 THE EQUATION IS BALANCED
2 HgO  2 Hg + O2
Element Reactant Side Product Side Hg O

21. General Guidelines:
The coefficients in a balanced equation are the lowest whole numbers that will balance the equation.
Useful to consider polyatomic ions as single entities, provided they maintain their identities in the reaction.
Subscripts are never changed!!!

22. Write and Balance the Equation
When gaseous butane (C4H10) is completely burned in oxygen, the products are carbon dioxide and water.

23. C4H10 (g) + O2 (g) → CO2(g) + H2O(l)
Balance the Equation
C4H10 (g) + O2 (g) → CO2(g) + H2O(l) 4
Reactant Side Product Side
C 4 1
H 10 2 O 4 9
Place a 4 in front of CO2 to balance C.
There are four C on the reactant side and there is one C on the product side.

24. C4H10 (g) + O2 (g) → CO2(g) + H2O(l)5
Reactant Side Product Side
C 4 4
H 10 2 O 10 13
Place a 5 in front of H2O to balance H.
There are 10 H on the reactant side and there are two H on the product side.

25. C4H10 (g) + O2 (g) → 4 CO2(g) + 5 H2O(l)8 10
Reactant Side Product Side
C 4 4
H 10 10 O 20 8 26
To balance O double all of the other coefficients.
There is no whole number coefficient that can be
placed in front of O2 to balance O.

26.  THE EQUATION IS BALANCED 
C4H10 (g) + O2 (g) → CO2(g) + H2O(l) 2 8 10 13
Reactant Side Product Side
C 8 8
H 20 20 O 26
Place a 13 in front of O2 to balance O.
There are now 26 O on the product side.

27. Balancing Example Al + CuCl2  Cu + AlCl3 Al Cu Cl 2 3 3 2 2  1 1 2 3
Aluminum and copper(II) chloride react to form copper and aluminum chloride.
Al CuCl2  Cu + AlCl3 Al Cu Cl 2 3 3 2
2 
 2
 6
3 
6 
 3

28. Exercise NH3 + O2 →NO + H2O 4NH3+5O2→4NO +6H2O C+ O2 → CO
2CO + O2 → 2CO2
CO + O2 → CO2
H2+Br2→HBr
H2 + Br2 → 2HBr
K +H2O → KOH + H2
2K + 2H2O → 2KOH + H2
Mg + O2 →Mgo
O3 → O2
4NH3+5O2 →4NO+6H2O 2.

29. Exercise H2O2 →H2O + O2 2H2O2 → 2H2O + O2 N2 + H2 →NH3 N2 + 3H2 → 2NH3
Zn + AgCl → ZnCl2 + Ag
Zn + 2AgCl → ZnCl2 + 2Ag
S8 +O2→ SO2
S8 + 8O2 →8SO2
NaOH + H2SO4 → Na2SO4 +H2O
2NaOH + H2SO4 → Na2SO4 + 2H2O
Cl2 + NaI → NaCl + I2
Cl2 + 2NaI → 2NaCl + I2
KOH +H3PO4 → K3PO4+ H2O
3KOH + H3PO4 → K3PO4 + 3H2O

30. Types of Chemical Reactions
Combustion Reactions - involve oxygen (typically from the air)
Synthesis

31. Two reactants combine to form one product.
Synthesis Reaction
Two reactants combine to form one product.
A + B  AB

32. In a synthesis reaction, there is only one product.
2Ca(s) + O2(g)  2CaO(s)
In a synthesis reaction, there is only one product.
4Al(s) + 3O2(g)  2Al2O3(s)
S(s) + O2(g)  SO2(g)
2Al(s) + 3Cl2(g)  2AlCl3(s)
Na2O(s) + H2O(l)  2NaOH(aq)

33. AB  A + B Decomposition Reaction
A single substance breaks down to give two or more different substances.
AB  A + B

34. In a decomposition reaction, there is only one reactant.
2Ag2O(s)  4Ag(s) + O2(g)
2H2O2(l)  2H2O(l) + O2(g)
2KClO3(s)  2KCl(s) + O2(g)
2NaNO3(s)  2NaNO2(s) + O2(g)

35. A + BC  AC + B Single Displacement Reaction
One element reacts with a compound to replace one of the elements of that compound.
A + BC  AC + B

36. Mg(s) + HCl(aq)  H2(g) + MgCl2(aq)
Notice that the formula for the magnesium-chloride compound is MgCl2, not MgCl. You will have to be able to write correct formulas based on your understanding of the ions involved; (Mg  Mg2+, therefore 2 Cl- ions are required to balance the charges).

37. The Activity Series
An activity series is a list of substances ranked in order of reactivity. The activity series is a useful guide for predicting of metal displacement reaction.
For example magnesium can knock hydrogen out of solution, so it is considered more reactivity than the element hydrogen.

38. Metals K Ca Na Mg Al Zn Fe Ni Sn Pb H Cu Ag Hg
The Activity Series
An atom of an element in the activity series will displace an atom of any element below it from one of its compounds .
Metals K Ca Na Mg Al Zn Fe Ni Sn Pb H Cu Ag Hg
Sodium (Na) will displace any element below it from one of its compounds.
increasing activity

39. Mg(s) + PbS(s)  MgS(s) + Pb(s)
Metals Mg Al Zn Fe Ni Sn Pb
Magnesium is above lead in the activity series.

40. Ag(s) + CuCl2(s)  no reaction?
Metals Pb H Cu Ag Hg
Silver is below copper in the activity series.

41. Cl2(g) + CaBr2(s)  CaCl2(aq) + Br2(aq)
There is also a halogen activity series – it has the same order as the halogens are found in the Periodic Table
Cl2(g) + CaBr2(s)  CaCl2(aq) + Br2(aq) ?
Halogens F2 Cl2 Br2 I2
Chlorine is above bromine in the activity series.

42. AB + CD  AD + CB Double Displacement Reaction A combines with D
The reaction can be thought of as an exchange of positive and negative groups.
Two compounds exchange partners with each other to produce two different compounds.
And B combines with C
AB + CD  AD + CB

43. One or more of the following will accompany a Double Displacement Reaction
formation of a precipitate
formation of a gas (release of bubbles)
release or absorption of heat
formation of a molecular compound (e.g., H2O)
A precipitate is a reaction product that isn’t soluble in the reaction medium; it falls out of solution as it forms.

44. Acid Base Neutralization
acid + base → salt + water
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(l)

45. Formation of an Insoluble Precipitate
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)

46. CaO(s) + 2HCl(aq)  CaCl2(s) + H2O(l)
Metal Oxide + Acid
metal oxide + acid → salt + water
CuO(s) + 2HNO3(aq)  Cu(NO3)2(aq) + H2O(l)
CaO(s) + 2HCl(aq)  CaCl2(s) + H2O(l)

47. indirect gas formation
Formation of a Gas
H2SO4(aq) + 2NaCN(aq)  Na2SO4(aq) + 2HCN(g)
NH4Cl(aq) + NaOH(aq)  NaCl(aq) + NH4OH(aq)
indirect gas formation
NH4OH(aq)  NH3(g) + H2O(l)

48. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Combustion Reaction
A reaction in which a substance reacts with oxygen and which proceeds with the evolution of heat.
2A + O2  2AO
Products:
contain oxygen
hydrocarbons (CxHy) form CO2 + H2O
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

49. Chemical Equations and the Mole
P4O H2O  4 H3PO4
1 molecule of P4O10 reacts with 6 molecules of H2O to give 4 molecules of H3PO4
1 mole of P4O10 reacts with 6 moles of H2O to give 4 moles of H3PO4

50. 6 Conversion Factors P4O10 + 6 H2O  4 H3PO4 1 mol P4O10 6 mol H2O
6 mol H2O 1 mol P4O10
4 mol H3PO4 1 mol P4O10
1 mol P4O10 4 mol H3PO4
Can you write two more?

51. Stoichiometry and Chemical Equations
The coefficients in a chemical equation not only describe the number of atoms/molecules being reacted/produced, they also describe the number of moles of atoms/molecules being reacted/produced.
N2(g) + 3H2(g)
2NH3(g)
Suppose 5.46 mol N2 completely react in this fashion:
5.46 mol N2
3 mol H2
1 mol N2
= 16.4 mol H2 are needed to react
5.46 mol N2
2 mol NH3
1 mol N2
= 10.9 mol NH3 must be produced

52. Moles A to Moles of B

53. Moles A to Moles B

54. The Limiting Reactant
For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others
When this reactant is used up, the reaction stops and no more product is made
The reactant that limits the amount of product is called the limiting reactant
sometimes called the limiting reagent
the limiting reactant gets completely consumed
Reactants not completely consumed are called excess reactants
The amount of product that can be made from the limiting reactant is called the theoretical yield

55. Theoretical, Actual & Percent Yields
Theoretical yield: the amount of product you would get if the reaction occurs with complete efficiency.
Actual yield: what you “actually” get when you perform the reaction.
Percent yield: the actual yield divided by the theoretical yield multiplied by 100.
% Yield = (Actual/Theoretical) x 100

56. Practice — How many moles of Si3N4 can be made from 1
Practice — How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the reaction 3 Si + 2 N2  Si3N4?
Given:
Find:
1.20 mol Si, 1.00 mol N2
mol Si3N4
Conceptual Plan:
Relationships:
2 mol N2 : 1 Si3N4; 3 mol Si : 1 Si3N4
mol Si
mol Si3N4
Pick least
amount
Limiting
reactant and
theoretical
yield
mol N2
mol Si3N4
Solution:
Limiting
reactant
Theoretical
yield 56

57. Theoretical, Actual & Percent Yields
Theoretical yield: the amount of product you would get if the reaction occurs with complete efficiency.
Actual yield: what you “actually” get when you perform the reaction.
Percent yield: the actual yield divided by the theoretical yield multiplied by 100.
% Yield = (Actual/Theoretical) x 100

58. Percent Yield
measured in lab
calculated on paper

59. When 45. 8 g of K2CO3 react with excess HCl, 46. 3 g of KCl are formed
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
? g
actual: 46.3 g

60. K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
? g
actual: 46.3 g
Theoretical Yield:
45.8 g
K2CO3
2 mol
KCl
1 mol
K2CO3
74.55
g KCl
1 mol
KCl
1 mol
K2CO3 g
= 49.4
g KCl

61. K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
46.3 g
49.4 g
 100 =
93.7%
% Yield =