4.2 STOICHIOMETRY

CHAPTER 4

Chemical Equation
& Stoichiometry

4.1 Chemical Equation

4.2 Stoichiometry

F2F: 13 HOURS NF2F: 13 HOURS

1

4.2
Stoichiometry

The study of quantitative aspects of
chemical formulas and reaction.

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Learning Outcomes

At the end of the lesson, students should be
able to:

(a) Stoichiometry mass calculation:
Calculate the amount of reactant and product from a balanced
chemical equation.
Calculation involve:
i. Reacting masses and moles
ii. Volume of gases at room condition and s.t.p (e.g combustion of
hydrocarbon
iii. Volume and concentration of solutions.

3

4

Stoichiometry Calculation
Four quantities in chemis5 try!!!

Interpreting chemical equation…

1. 2Na (s) + Cl2 (g)  2NaCl (s)

2 mol sodium reacted with 1 mol chlorine producing 2 mol sodium
chloride

2 (23.0 ) g Na reacted with 35.5(2) g Cl2 producing 2 (23.0 + 35.5) g NaCl
2 (6.02 x 1023) atoms Na reacted with 6.02 x 1023 molecules Cl2 producing

22.4 dm3 Cl2 (at STP) or 2 (6.02 x 1023) units NaCl

6

Interpreting chemical equation…

2. 4Fe (s) + 3O2 (g)  2Fe2O3 (s)

4 mol iron _______3 mol oxygen ______2 mol iron(III) oxide

4 (55.9)g Fe ≡ 3 (16.0x2)g O2 ≡ 2 [(55.9x2) + 16.0x3] g Fe2O3
4 (6.02 x 1023) atoms Fe ≡ 3 (6.02 x 1023) molecules O2 ≡
3 (22.4 dm3) O2 (at STP) ≡ 2 (6.02 x 1023) units Fe2O3 7

Interpreting chemical equation…

3. CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)

? mol methane ______? mol oxygen ______? mol carbon dioxide
_________ ? mol water

? g CH4 ≡ ? g O2 ≡ ? g CO2 ≡ ? g H2O

? molecules CH4 ≡ ? molecules O2 ≡

? dm3 O2 (at STP) ≡ ? molecules CO2 ≡ ? dm3 CO2 (at STP) ≡

? molecules H2O 8

Exercise

Interpret quantitatively the chemical equation below:
CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)

Solution

1 mol ≡ 2 moles ≡ 1 mol ≡ 1 mol ≡ 1 mol

100.1 g ≡ 2(36.5) g ≡ 111.1 g ≡ 44.0 g ≡ 18.0 g

6.02x1023 2(6.02x1023 ) 6.02x1023 6.02x1023 6.02x91023
molecules ≡ molecules ≡
molecules ≡ molecules ≡ molecules

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Example 1

How many moles of water, H2O will be formed when
0.25 moles of ethanol, C2H5OH burns in oxygen, O2?

Step 1: Write the balance chemical equation.
C2H5OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (l)

Step 2: Interpret the equation / compare the stoichiometric
relationship between C2H5OH (mole) and H2O (mole) & solve
the problem.

From the equation, 1 mol C2H5OH ≡ 3 mol H2O

 0.25 mole C2H5OH ≡ 0.25 mol C2H5OH x 3 mol H2O

1 mol C2H5OH 10
= 0.75 mol H2O

Example 2

Potassium sulphate, K2SO4 and water, H2O are produced from the
neutralisation reaction of sulphuric acid, H2SO4 and potassium
hydroxide, KOH.

(a) Write a balanced equation for the neutralisation reaction.

(b) Calculate the mass of sulphuric acid required to produce 215 g
of potassium sulphate.

(c) Calculate the number particles of potassium sulphate produced
if 1.23 mol of potassium hydroxide is used.

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Solution

(a) H2SO4 (aq) + 2KOH (aq)  K2SO4 (aq) + 2H2O (l)
? g 215 g

(b) 1 mole H2SO4 ≡ ((2x1)+(32.1)+(16x4)) = 98.1 g
1 mole K2SO4 ≡ ((2x39.1)+(32.1)+(16x4)) = 174.3 g

From the equation,

1 mole H2SO4 ≡ 1 mole K2SO4
98.1 g H2SO4 ≡ 174.3 g K2SO4
∴ 215 g K2SO4 ≡ 215 g K2SO4 x 98.1 g H2SO4

174.3 g K2SO4 12
= 121 g H2SO4

Solution

(c) H2SO4 (aq) + 2KOH (aq)  K2SO4 (aq) + 2H2O (l)

1.23 mol particles?

From the equation,

2 mol KOH ≡ 1 mole K2SO4 ≡ 6.02x1023 particles K2SO4

 1.23 mol KOH ≡ 1.23 mol KOH x 6.02x1023 particles K2SO4
2 mol KOH

= 3.70 x 1023 particles (molecules) K2SO4

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Learning Outcomes

At the end of the lesson, students should be
able to:

(b) Define:
i. limiting reactant
ii. theoretical yield
iii. actual yield; and
iii. Percentage yield.

14

Learning Outcomes

At the end of the lesson, students should be
able to:

(c) Perform calculation involving limiting reactant and
percentage yield.

15

16

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The analogy…

In theory, to make a bicycle, we need:

+

1 body 2 tyres 1 bicycle

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In a certain situation, suddenly we have:

+

3 bodies 4 tyres 2 bicycles

1. Which ‘reactant’ is in excess (berlebihan)? [so called : excess reactant]
2. Which ‘reactant’ is just enough (ngam-ngam)? [so called : limiting reactant]

3. Which ‘reactant’ limits the quantity of the ‘products’ / bicycle form?
19

In another situation, we have:

+

2 bodies 6 tyres 2 bicycles

1. Which ‘reactant’ is in excess (berlebihan)? [so called : excess reactant]
2. Which ‘reactant’ is just enough (ngam-ngam)? [so called : limiting reactant]

3. Which ‘reactant’ limits the quantity of the ‘products’ / bicycle form?
20

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Definition Limiting
reactant is the
Excess reactant reactant that is
is the reactant
completely
that is not consumed in a
completely reaction and
consumed in a
reaction and limits the
remains at the amount of
end of the products
reaction.
formed.

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In the next slides, we will
relate the previous
analogy of limiting

reactant with the ‘real’
chemical reaction.

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EXAMPLE 1

The chemical equation for formation of
iron (II) sulfide is:

Fe (s) + S (s)  FeS (s)

1 mol 1 mol 1 mol

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In a certain reaction, we have:

Fe (s) + S (s)  FeS (s)

4 mol 6 mol 4 mol

a. Which reactant is the excess reactant?

Sulphur, S

b. Which reactant is the limiting reactant?

Iron, Fe 24

So how did you determine
the excess reactant & the

limiting reactant?

Did you compare the
number of mole of
the reactant Fe with number
of mole of the reactant S?

25

Determining Limiting Reactant

Comparing Mole Reactant, A with Mole Reactant, B.

STEP 1 STEP 2 STEP 3 STEP 4

Write the Compare mole
chemical
equation. Calculate Calculate available and

A mole mole required of both

(available) (required) of reactants.

of both both

reactants. reactants. If n required > navailable
⸫ Limiting reactant

BALANCED If n available > n required
equation!! ⸫ Excess reactant

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EXAMPLE 1

Fe (s) + S (s)  FeS (s)

Given, 4 mol 6 mol 4 mol

Determine the limiting reactant.

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Step 1 – 4: Determining Limiting reactant

(a) Step 1 Fe (s) + S (s)  FeS (s)

Write the
balanced
chemical
equation.

(b) Step 2 given:
4 mol Fe

Calculate mole
(available) of

both reactants. @

6 mol S 28

Step 3 From balanced equation:

Calculate mole

(required) of 1 mol Fe ≡ 1 mol S
both reactants.
4 mol Fe ≡ 4 mol S

@

1 mol S ≡ 1 mol Fe
 6 mol S ≡ 6 mol Fe

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Step 4

Compare mole

available and Since mole Fe available less

required of both than mole Fe required,

reactants. Fe is in limits.

 Fe is the limiting reactant.

If n required > n @
⸫avaLilaibmleiting

reactant Since mole S available more

than mole S required, S is

If n available > n excess.
⸫reqEuirxecd ess  Fe is the limiting reactant.

reactant 30

EXAMPLE 2
The reaction between iron, Fe and sulphur, S is
given as :

Fe (s) + S (s)  FeS (s)

If 23.54 g of iron is reacted with 12.0 g of
sulphur, determine the limiting reactant.

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Step 1 – 4: Determining Limiting reactant

(a) Step 1 Fe (s) + S (s)  FeS (s)

Write the
balanced
chemical
equation.

(b) Step 2 Mol of Fe
Calculate mole = mass/molar mass
(available) of = 23.54 / 55.9
both reactants. = 0.4211 mol Fe

Mol of S 32
= mass/molar mass
= 12.0 / 32.1
= 0.3738 mol S

Step 3 From balanced equation:

Calculate mole 1 mol Fe ≡ 1 mol S
(required) of 0.4211 mol Fe ≡ 0.4211 mol S
both reactants.

@
1 mol S ≡ 1 mol Fe
 0.3738 mol S ≡ 0.3738 mol Fe

33

Step 4

Compare mole

available and Since mole S available less than mole

required of both S required,

reactants. S is in limits.

 S is the limiting reactant.

If n required > n available @
⸫ Limiting reactant

If n available > n required Since mole Fe available more than
⸫ Excess reactant mole Fe required, Fe is excess.

 S is the limiting reactant.

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SOLUTION

Fe (s) + S (s)  FeS (s)

23.54 g 12.0 g

STEP 2 STEP 3 STEP 4
Mol S required= 0.4211 mol
Mol of Fe From the balanced equation: Mol S available =0.3738 mol
Mol S required > mol S
= mass/molar mass 1 mol Fe ≡ 1 mol S available
S = limiting reactant.
= 23.54 / 55. 9 0.4211 mol Fe ≡ 0.4211 mol S
35
= 0.4211 mol Fe

Mol of S
= mass/molar mass
= 12.0 / 32.1
= 0.3738 mol S

EXAMPLE 3

The reaction between iron and water is given as:
3Fe (s) + 4H2O (g)  Fe3O4 (s) + 4H2 (g)
If 23.54 g of iron is reacted with 15.0 g of water,
determine the limiting reactant in the reaction.

36

Step 1 – 4: Determining Limiting reactant

(a) Step 1 3Fe (s) + 4H2O (g)  Fe3O4 (s) +
4H2 (g)
Write the
balanced
chemical
equation.

(b) Step 2 nH2O available
Calculate mole = mass/ molar mass
(available) of = 15.0 / 18.0
both reactants. = 0.8333 mol

nFe available 37
= mass/ molar mass
= 23.54 / 55. 9
= 0.4211 mol

Step 3 From balanced equation:

Calculate mole 3 mol Fe ≡ 4 mol H2O
(required) of 0.4211 mol Fe ≡ 0.5615 mol H2O
both reactants. (mole required)

@

4 mol H2O ≡ 3 mol Fe
0.8333 mol H2O ≡ 0.6250 mol Fe
(mole required)

38

Step 4

Compare mole Since mole H2O available more than
available and mole H2O required, H2O is excess.
required of both  Fe is the limiting reactant.
reactants.
@
If n required > n
a⸫vaLilaibmleiting Since mole Fe available less than
reactant mole Fe required,
Fe is in limits.
If n available > n  Fe is the limiting reactant.
r⸫eqEuirxecd ess
reactant 39

EXAMPLE 4

When 2.38 g of magnesium is added to a 25 cm3 of 2.27
M hydrochloric acid, hydrogen gas is released.

(a) Determine the limiting reactant.
(b) Calculate the volume of hydrogen gas produced at

room temperature.
(c) Calculate the mass of excess reactant remained at

the end of the reaction.

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SOLUTION

Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
2.38 g 25 cm3, 2.27M

(a) Determine the limiting reactant

STEP 2 STEP 3 STEP 4

Mol of Mg From the equation: Mol HCl required
= mass/molar mass 1 mol Mg ≡ 2 mol HCl > mol HCl
= 2.38g/24.3gmol-1 0.0979 mol Mg ≡ 0.0979 mol Mg x 2 mol HCl available
= 0.0979 mol Mg
1 mol Mg HCl = limiting
= 0.196 mol HCl reactant.

Mol of HCl
= MV
= 2.27 (25/1000)
= 0.0567 mol HCl

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(b) Calculate the volume of hydrogen gas produced at
room temperature

SOLUTION

Since HCl is the limiting reactant, the amount of H2 gas is
depends on the amount of HCl used in the reaction.

Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)

0.0567 mol

From the chemical equation,

2 mol HCl ≡ 1 mol H2 ≡ 24 dm3 H2 (at Room T)
0.0567 mol HCl ≡ 0.0567 mol HCl x 24 dm3 H2
2 mol HCl

= 0.680 dm3 H2
42

(c) Calculate the mass of excess reactant remained at
the end of the reaction.

SOLUTION

Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)

0.0979 mol
(available)

Since HCl is the limiting reactant, then the excess reactant is Mg.

From the chemical equation,
2 mol HCl ≡ 1 mol Mg

 0.0567mol HCl ≡ 0.0567 mol HCl x 1 mol Mg
2 mol HCl

= 0.0284 mol Mg (mol of Mg required)
Mol of Mg remained after reaction = mol Mg available – mol Mg required

= 0.0979 mol – 0.0284 mol
= 0.0695 mol Mg
Mass of Mg remained = mol x molar mass = 0.0695 mo43l X 24.3 g mol-1

= 1.69 g Mg

Why is percentage yield/product important
(eg. in chemical industry)???44

The percentage yield is the ratio of the actual yield
(obtained from experiment) to the theoretical yield
(obtained from the chemical equation – from
calculation) multiply by 100%.

From experiment

From chemical equation

* Can be in : mass or mol or volume or number of

particles!!! 45

Theoretical Yield

Theoretical yield is the quantity of a product
obtained from the complete conversion of the
limiting reactant in a chemical reaction.

Actual Yield

Actual yield is the amount actually produced of
a product.

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EXAMPLE

In a certain experiment, 14.6 g of SbF3 was allowed
to react with excess CCl4. After the reaction was
over, 8.62 g of CCl2F2 was obtained. The equation
for the reaction is given as below:

3 CCl4 + 2 SbF3 3 CCl2F2 + 2 SbCl3

What was the percentage yield of CCl2F2?

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SOLUTION

Excess Limiting
reactant reactant

3 CCl4 + 2SbF3  3CCl2F2 + 2SbCl3

14.6 g Actual yield
8.62g

Percentage yield = Actual yield X 100 %
Theoretical yield

* To calculate the percentage yield, we have to first find
the theoretical yield (*actual yield is given).

48

Theoretical yield = Amount of product obtained from chemical
equation.

3 CCl4 + 2 SbF3  3 CCl2F2 + 2 SbCl3

From the equation,

2 mol SbF3 ≡ 3 mol CCl2F2

2 [122 + (3x19)] g SbF3 ≡ 3 [12 + (2x35.5) + (2x19)] g CCl2F2

 358 g SbF3 ≡ 360 g CCl2F2

14.6 g SbF3 ≡ 14.6 g SbF3 x 360 g CCl2F2

358 g SbF3

= 14.68 g CCl2F2 (theoretical yield)
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Percentage yield = Actual yield X 100 %
Theoretical yield

Percentage yield = 8.62 g x 100 %
14.68 g

= 58.72 %

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