This document contains notes from a chemistry class. It begins with an introduction and prayer. It then discusses various chemistry topics like balancing chemical equations, types of chemical reactions, limiting and excess reactants, and percent yield. It provides examples and explanations for each topic. Key points covered include the steps to balance equations, the characteristics of different reaction types (synthesis, decomposition, etc.), how to determine the limiting and excess reactants, and how to calculate theoretical and percent yields of a reaction. The document appears to be notes that a teacher provided to students for a general chemistry course.

1. HELLO,
Dear Students!
Ma’am Jhanel S. Punongbayan
SCIENCE 9A
GENERAL CHEMISTRY

2. PRAYER

3. ATTENDANCE

6. Chemy wants a cheese burger for
snacks, he asked his family if they
want cheese burger too and they all
said yes. They are 4 in the family.
Upon checking the refrigerator,
Chemy noticed that they only have 2
patties left and 1 bun. He decided to
go to the Supermarket near their
house to get other ingredients he’ll
be needing for the cheese burger.

8. = 3
= 4
= 4
= 4
= 2

9. 4 4 4 4
4

10. STOICHIOMETR
Y 2
BALANCING CHEMICAL EQUATIONS
AND
TYPES OF CHEMICAL REACTIONS

11. Write equations for
chemical reactions
and balance the
equations
01
Calculate percent
yield and
theoretical yield of
the reaction
02
Explain the concept
of limiting reagent in
a chemical reaction;
identify the excess
reagent(s)
03
Determine mass
relationship in a
chemical reaction
04

12. CHEMICAL
EQUATIONS
STEPS IN BALANCING
01

13. CHEMICAL EQUATION
A chemical equation is made up of
two parts:
the reactant and the product.
ingredie
nts
sandwich

14. CHEMICAL EQUATION
REACTANTS PRODUCT
R R PR
combine/deplete
to generate a new
material
the new material
that results from the
reaction of the
reactants

15. Fe2O3 + 3 H2SO4 ⟶ Fe2(SO4)3 + 3
H2O
REACTANTS
The substances undergoing
reaction
their formulas are placed on
the left side of the equation

16. Fe2O3 + 3 H2SO4 ⟶ Fe2(SO4)3 + 3
H2O
PRODUCTS
substances generated
by the reaction
their formulas are placed
on the right side of the
equation.

17. Fe2O3 + 3 H2SO4 ⟶ Fe2(SO4)3 + 3
H2O
an arrow (⟶) separates
the reactant and product
(left and right) sides of the
equation.
Plus signs (+) separate
individual reactant and
product formulas

18. Fe2O3 + 3 H2SO4 ⟶ Fe2(SO4)3 + 3
H2O
Coefficients
relative numbers of
reactant and product species
Numbers placed to the
left of each formula
A coefficient of 1 is
typically omitted.

19. BALANCING
CHEMICAL
EQUATION

20. Let us have the reaction in the synthesis of
water:
In this chemical reaction, it can be seen that the
hydrogen gas and the oxygen gas react to form a new
substance, water.
H2 + O2 ⟶
H2O
Note that hydrogen and oxygen, and other nonmetals do not exist in
elemental form unlike other elements. They are usually reacting as a
diatomic molecule.

21. H2 + O2 ⟶
H2O
According to the Law of Conservation of Matter, the
amount of matter in any system that is closed to the
transfer of matter remains constant.
"what comes in, comes out"
It can be observed in the example above that there are two hydrogen
atoms that will react with the two oxygen atoms. Following the
reaction, the product (water) includes two hydrogen atoms but only one
oxygen atom. This is a concern because the law of conservation of
matter applies to all chemical reactions.

22. H2 + O2 ⟶ H2O
H2 + O2 ⟶ 2H2O
2H2 + O2 ⟶ 2H2O
Count the number of atoms in each element on the reactant and
product sides and compare them to see if the chemical equation is
balanced.
Take note that we can use any coefficient available as long as it is the least
possible number that can balance a chemical reaction.

23. Fe2O3 + C ⟶ Fe + CO2
This shows the reaction of Iron (III) oxide and carbon to form iron and
carbon dioxide.
Take note that balancing a chemical equation is a trial-and-error process.
2 Fe2O3 + C ⟶ Fe + 3
CO2
Let us balance first Oxygen having the largest number of atoms in both sides of the equation.

24. 2 Fe2O3 + 3 C ⟶ 4 Fe + 3 CO2
Let us try to balance the rest of the equation.
WE BALANCED THE
CHEMICAL EQUATION!

25. LET’S TRY
THESE:
1. ____ Sb + ____ Cl₂ → _____ SbCl₃
2. ____ N₂ + ____ H₂ → _____ NH₃
2 Sb + 3 Cl₂ → 2 SbCl₃
N₂ + 3 H₂ → 2 NH₃

26. TYPES OF
CHEMICAL
REACTIONS

27. 1. Synthesis
Reactions
2.
Decomposition
3. Single
Replacement
4. Double
Replacement
5. Combustion

28. SYNTHESIS
REACTIONS
- two or more reactants combine to make one
type of product. The general equation for
synthesis reaction is A + B → AB. It’s like:
simpler elements or molecules
more complex
molecule

29. 2H₂ + O₂ →
2H₂O
two elements (hydrogen and oxygen) are combining to
form one product (water)
You should be able to write the chemical equation for a
synthesis reaction if you are given a product by picking out its
elements and writing the equation.
Also, if you are given elemental reactants and told that the
reaction is a synthesis reaction, you should be able to predict
the products.

30. DECOMPOSITION
REACTIONS
- one type of reactant breaks down to form two or more
products
- for all reactions of this type, there is only one reactant.
The general equation for decomposition reaction is
AB → A + B. It’s like:

31. Look at the formula for magnesium nitride, Mg3N2.
What elements do you see in this formula? Now we
can write a decomposition reaction for magnesium
nitride. Notice there is only one reactant.
Mg3N2 → 3Mg +
N2

32. SINGLE REPLACEMENT
REACTIONS
- one element reacts with one compound to form products.
- The single element is said to replace an element in the
compound when products form, hence the name single
replacement.
- Metal elements will always replace other metals in ionic
compounds or hydrogen in an acid.
- Nonmetal elements will always replace another nonmetal in
an ionic compound.

33. Zn + Cu(NO3)2 → Zn(NO3)2 + Cu
Consider this example.
Notice that the metal element, Zn, replaced the metal in
the compound Cu(NO3)2. A metal element will always replace
a metal in an ionic compound. Also, note that the charges of
the ionic compounds must equal zero.
To correctly predict the formula of the ionic product, you
must know the charges of the ions you are combining.

34. DOUBLE
REPLACEMENT
— also called double displacement, exchange, or metathesis
- occurs when parts of two ionic compounds are exchanged,
making two new compounds.
PRECIPITATIO
N
NEUTRALIZA
TION
two aqueous compound reactants
combine to form products where
one of the products is an
insoluble solid
two reactant compounds are an
acid and a base, and the two
products are a salt and water
(acid + base → salt + water).

35. AgNO3(aq) + NaCl(aq) → AgCl(s) +
NaNO3(aq)
For example, when solutions of silver nitrate and
sodium chloride are mixed, the following reaction
occurs:
This is an example of a precipitation reaction.
Notice that two aqueous reactants form one solid,
the precipitate, and another aqueous product.

36. 2NaOH(aq) + H2SO4(aq) →
Na2SO4(aq) + 2H2O(l)
An example of a neutralization reaction occurs when
sodium hydroxide, a base, is mixed with sulfuric acid:

37. COMBUSTION
REACTION
a reaction in which a substance reacts with oxygen
gas, releasing energy in the form of light and heat.
- must involve O₂ as one reactant.
In other words, the only part that changes from
one combustion reaction to the next is the actual
hydrocarbon that burns.

38. 2C8H18 + 25O2 → 16CO2 +
18H2O
Look at the reaction for the combustion of octane, C8H18,
below. Octane has 8 carbon atoms hence the prefix “oct”.
This reaction is referred to as complete combustion.
Complete combustion reactions occur when there is enough
oxygen to burn the entire hydrocarbon. This is why there
are only carbon dioxide and water as products.

39. HELLO
GRADE 9!
Student Teacher
Ma’am Jhanel S. Punongbayan
SCIENCE 9A
GENERAL CHEMISTRY

40. LIMITING AND
EXCESS
REACTANT

41. LIMITING
REACTANTS
- those that completely react with the other substance.
- In simple terms, everything of that specific reactant is
used or reacts with the other substance to form new
substances or the products.
EXCESS
REACTANTS
reactant did not react completely.

42. Let us take a look at this analogy of the limiting and excess
reactant: To make a simple cheese sandwich, we need two loaves
of bread and a slice of cheese. We can see that the ratio of
bread to cheese is 2:1. If you have 28 loaves of bread and 11
slices of cheese, we will be able to use all the slices of cheese,
but not all the loaves of bread. 6 loaves of bread will remain
because the cheese is not sufficient enough to use all the
available loaves of bread.

44. How are we going to determine which of the reactants in a
chemical reaction is present in excess and exact amounts?
TWO
WAYS
The first way to solve the amount of product that will be produced
if both of the reactants will be completely used. The reactant which
produced the lesser amount of product is the limiting reactant,
while the other is the excess reactant.
Another way is to compare directly the two reactants and solve for
the amount of the reactant that will be used assuming that the
other reactant will be used completely.
Depending on the result, we can conclude which of
the two is the limiting and the excess reactant.

45. The reaction of iron (III) oxide and carbon produces iron and
carbon dioxide. Suppose there are 3 moles of each of the
reactants which are available to react. Determine the limiting
and the excess reactant. The chemical equation is shown
below:
Please take note that the concepts on balancing chemical
equations, molar/formula mass, and the mass-mole
relationships are required to be applied in these kinds of
problem.
2 Fe2O3 + 3 C → 4 Fe + 3
CO2
For the first solution, let us try to find the amount of
Fe that will be produced if both of the reactants will
be completely used.

46. 𝑚𝑜𝑙 𝐹𝑒 = 3 𝑚𝑜𝑙 𝐹𝑒2𝑂3 ×
𝟒 𝒎𝒐𝒍 𝑭𝒆
𝟑 𝒎𝒐𝒍 𝑭𝒆𝟐
𝑶𝟑
= 6 𝒎𝒐𝒍 𝑭e
𝑚𝑜𝑙 𝐹𝑒 = 3 𝑚𝑜𝑙 C ×
𝟒 𝒎𝒐𝒍 𝑭𝒆
𝟑 𝒎𝒐𝒍 𝑪
= 4 𝒎𝒐𝒍 𝑭e
Based on the results, if there are 3 moles of C present in the
reaction, only 4 moles of Fe will be produced as compared
with the 6 moles of Fe that will be produced if we will use all
the available amount of Fe2O3. This clearly shows that C is
our limiting, while the Fe2O3 is our excess reactant.

47. Let us try to find out the amount of Fe2O3 that will
react with 3 moles of carbon.
Solution 2.1:
The solution above simply means that it only requires 2
moles of Fe2O3 to completely use the available amount
of carbon, therefore, the available amount of Fe2O3 is
present in excess.

48. In case you want to solve for the amount of C that will react
with 3 moles of Fe2O3, the result will be just like this:
Solution 2.2:
The solution above shows that it will require 4.5 moles of C
to completely use 3 moles of Fe2O3.There is a problem here
because as stated in the problem, there are only 3 moles of
C available. This proves that C is our limiting reactant, while
Fe2O3 is our excess reactant.

49. Sample Problem 2:
The combustion of any hydrocarbon produces water
and carbon dioxide. If there are 30 grams of C4H10
and 45 grams of O2 in a container, determine which is
the excess and the limiting reactant, and solve for
the amount (in grams) of the excess reagent. Refer to
the balanced chemical equation below:
Solution 1:
𝑔 𝐶4𝐻10= 45 𝑔 𝑂 ×
𝟏 𝒎𝒐𝒍 𝑶𝟐
𝟑𝟐 𝒈 𝑶𝟐
x
𝟐 𝒎𝒐𝒍 𝑪𝟒𝑯𝟏𝟎
𝟏𝟑 𝒎𝒐𝒍 𝑶𝟐
x
𝟓𝟏.𝟏𝟒 𝒈 𝑪𝟒𝑯𝟏𝟎
𝟏 𝒎𝒐𝒍 𝑪𝟒
𝑯𝟏𝟎
= 12.58 g C4H10
Based on the solution, only 12.58 grams of C4H10 can be burned by
45 grams of O2. This means that O2 is our limiting reactant, while
C4H10 is our excess reactant.

50. To determine the amount of excess reactant,
subtract the answer from the available amount
of excess reactant.
𝐸𝑥𝑐𝑒𝑠𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 = 30 𝑔 𝐶4𝐻10 − 12.58 𝑔 𝐶4𝐻10
= 𝟏𝟕. 𝟒𝟐 𝒈 𝑪𝟒𝑯𝟏𝟎

51. PERCENT
YIELD

52. 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑌𝑖𝑒𝑙𝑑
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑
x 100%
the maximum amount of product that
can be formed from a certain reaction
using stoichiometric calculations
the amount of product formed from
the reaction carried out in a laboratory
the ratio of the theoretical yield to
the actual yield, which is expressed
in percentage.

53. 2 KClO3 → 2 KCl + 3 O2
After conducting an
experiment, 14.5 grams of O2
was produced by the
decomposition of 40 grams of
KClO3. Solve for the
theoretical yield and percent
yield of the reaction. Refer
to the chemical equation
below:
Sample Problem 1

54. SOLUTION
To solve for the percent yield, we must first solve for
theoretical yield. Using stoichiometric calculation, let us
determine the amount of O2 that will be produced from 40 grams
of KClO3. Take note that we will be using the mass – mass
conversions in this problem.
Based on the solution, the theoretical yield is 15.67 grams of O2.
From here, we can now solve for the percent yield.

56. Commonly, percent yields are reasonable less than
100% because of some unavoidable factors.
However, there can have some errors in percent
yield. Due to impurities in the products, percent
yield can sometimes be more than 100%. This is the
reason why chemists are very careful in carrying out
a chemical reaction to assure the purity of the
products.

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