1.0 Matter

2. Add OH- at both side, equal to the number of H+

3Cu+2NO3- + 8H+ +8OH- 3Cu2++2NO +4H2O+ 8OH-

3. Combine (OH-) and (H+) to form H2O

3Cu + 2NO3- + 8H2O 3Cu2+ + 2NO + 4H2O + 8OH-

101

4. Cancel any H2O that are equal at both side

3Cu + 2NO3- + 8H2O 3Cu2+ + 2NO + 4H2O + 8OH-
3Cu + 2NO3- + 4H2O 3Cu2+ + 2NO + 8OH-

102

EXERCISE 26

Balance the following redox equations:
a. In Acidic Solution

i. Cu + NO3- + H+ → Cu2+ + NO2 + H2O
ii. MnO4- + H2SO3 → Mn2+ + SO42- + H2O + H+

b. In Basic Solution
i. ClO- + S2O32- → Cl- + SO42-
ii. Cl2 → ClO3- + Cl-

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ANSWER OF EXERCISE 26

a. In Acidic Solution
i. Cu + 2NO3- + 4H+ → Cu2+ + 2NO2 + 2H2O
ii. 2MnO4- + 5H2SO3 → 2Mn2+ + 5SO42- + 3H2O + 4H+

b. In Basic Solution
i. 4ClO- + S2O32- + 2OH- → 4Cl- + 2SO42- + H2O
ii. 3Cl2 + 6OH- → ClO3- + 5Cl- + 3H2O

104

STOICHIOMETRY

Study of reactants and products in a chemical reaction.
Eg:
CaCO3(s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)
1 mole of CaCO3 reacts with 2 moles of HCl to yield
1 mole of CaCl2, 1 mole of CO2 and 1 mole of H2O.

1 mole CaCO3 Ξ 2 moles HCl Ξ 1 mole CaCl2
Ξ 1 moles CO2 Ξ 1 mole H2O

105

QUESTION 27

How many moles of hydrochloric acid, HCl do we need to react
with 0.5 moles of zinc?

Zn + 2 HCl ZnCl2 + H2

1 mol Zn Ξ 2 mol HCl

0.5 mol Zn Ξ 0.5 mol Zn × 2 mol HCl
1 mol Zn

= 1.0 mol HCl

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LIMITING REACTANT

LIMITING EXCESS
REACTANT REACTANT

Reactant that is Reactant that is not
completely consumed completely consumed

in a reaction and in a reaction and
limits the amount of remains at the end of

products formed. the reaction.

107

Reactant Excess reactant

+

Limiting reactant

?

DECIDING THE LIMITING AND
EXCESS REACTANT

Find the limiting reactant and mass of excess

reactant remain when 1.5 g of CaCO3 react with
0.73 g of HCl.

a) Write the balanced equation.

CaCO3 + 2HCl CaCl2 + CO2 + H2O

109

b) Calculate the given moles of each reactant
in the reaction.

n CaCO3 = 1.5 g n HCl = 0.73 g .
(given) 100.1 g/mol (given) 36.5 g/mol

= 0.014985 mol = 0.02 mol

c) Use the balanced equation to determine the mole
ratio of reactant used

From equation :

1 mol CaCO3  2 mol HCl

110

d) Compare given moles of reactants to mole ratio of
balance equation.

1 mol CaCO3  2 mol HCl

0.014985 mol CaCO3  0.014985 mol x 2 mol HCl
1 mol

= 0.02997 mol HCl (needed)

nHCl given (0.02 mol) is less than nHCl needed (0.02997
mol)

 HCl is limiting reactant

111

e) Limiting reactant – comletely consumed

Excess reactant - left after reaction complete

CaCO3 is excess reactant.
Based on the balanced equation,

2 mol HCl Ξ 1 mol CaCO3

0.02 mol HCl Ξ 0.02 mol x 1 mol CaCO3
2 mol

= 0.01 mol CaCO3 (reacted)

Mole CaCO3 remain = n of CaCO3 given – n of CaCO3 reacted
= 0.014985 mol – 0.01 mol
= 0.004985 mol

Mass of CaCO3 remain = n x molar mass
= 0.004985 mol x 100.1 g/mol
= 0.499 g

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QUESTION 28

The reaction between aluminium and iron (III) oxide can
generate temperature till 3000 0C and used in welding

metal:

2Al + Fe2O3 Al2O3 + 2 Fe

In a process, 124 g of Al are reacted with 601 g of Fe2O3

i) Calculate the mass of Al2O3 formed.
ii) How much of the excess reagent is left at the end of

reaction ? (Ans: 234.22 g, 234.06 g)

113

i. n Al = 124 g n Fe2O3 = 601 g .
(given) 27.0 g/mol (given) 159.8 g/mol

= 4.5926 mol = 3.7610 mol

Base on balanced equation,

2 mol Al Ξ 1 mol Fe2O3
4.5926 mol Al
Ξ 4.5926 mol × 1 mol Fe2O3
2 mol
= 2.2963 mol Fe2O3 (needed)

Since mol Fe2O3 given (3.7610 mol) is more than mol Fe2O3
needed (2.2963 mol), thus Fe2O3 is excess reactant.
 Al is limiting reactant.

Base on balanced equation,

2 mol Al Ξ 1 mol Al2O3
4.5926 mol
4.5926 mol Al Ξ 2 mol x 1 mol Al2O3

= 2.2963 mol Al2O3

 mass of Al2O3 formed = 2.2963 mol x 102 g/mol
= 234.22 g

ii. Base on balanced equation,

2 mol Al Ξ 1 mol Fe2O3

4.5926 mol Al Ξ 4.5926 mol ×1 mol Fe2O3
2 mol

= 2.2963 mol Fe2O3 (reacted)

n of Fe2O3 remain = n of Fe2O3 given – n of Fe2O3 reacted
= 3.7610 mol – 2.2963 mol

= 1.4647 mol

mass of Fe2O3 remain = n x molar mass
= 1.4647 mol x 159.8 g/mol

= 234.06 g

PERCENTAGE YIELD

Ratio of the actual yield (obtained from experiment)
to the theoretical yield (obtained from stoichiometry

calculation) multiply by 100%

Percentage yield = actual yield x 100 %

theoretical yield

116

QUESTION 29

In a certain experiment, 14.6 g of SbF3 was allowed to
react with CCl4 in excess. After the reaction was finished,
8.62 g of CCl2F2 was obtained. [ Ar Sb = 122, F = 19, C= 12, Cl = 35.5 ]

3CCl4 + 2SbF3 3CCl2F2 + 2SbCl3

a) What was the theoretical yield of CCl2F2 in grams ?
b)What was the percentage yield of CCl2F2 ?

(Ans: 14.80 g, 58.24 %)

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a) SbF3 is limiting reactant.
n of SbF3 = 14.6 g = 8.1564 x 10-2 mol
179 g/mol

Base on balanced equation,

2 mol SbF3 Ξ 3 mol CCl2F2
8.1564 × 10−2 mol
8.1564 x 10-2 mol SbF3 Ξ 2 mol × 3 mol CCl2F2

= 0.12235 mol CCl2F2

mass of CCl2F2 (theoretical yield) = n x molar mass
= 0.12235 mol x 121 g/mol

= 14.80 g

b) % yield of CCl2F2 = actual yield x 100 %
theoretical yield

= 8.62 g x 100 %

14.80 g

= 58.24 %

QUESTION 30
A student reacts benzene, C6H6 with bromine, Br2 as to prepare
bromobenzene, C6H5Br. What is the theoretical yield of
bromobenzene in this reaction when 30.0 g of benzene reacts
with 65.0 g of Br2? What was the percentage yield if the actual
yield of bromobenzene was 56.7 g? ( Ans: 60.34g, 93.97 %)

119

C6H6 + Br2 C6H5Br + HBr

n C6H6 = 30 g = 0.3846 mol n Br2 = 65.0 g = 0.4068 mol
(given) 78 g/mol (given) 159.8 g/mol

From equation, 1 mol C6H6 Ξ 1 mol Br2

0.3846 mol C6H6 Ξ 0.3846 mol Br2 (needed)

since mol Br2 given(0.4068 mol) is more than mol Br2 needed
(0.3846mol), thus Br2 is the excess reactant.

 C6H6 is the limiting reactant.

1 mol C6H6 Ξ 1 mol C6H5Br
0.3846 mol C6H6 Ξ 0.3846 mol C6H5Br

mass of C6H5Br (theoretical yield) = 0.3846 mol x 156.9 g/mol
= 60.34 g

% yield of C6H5Br = 56.7 g x 100 %
60.34 g

= 93.97 %

EXERCISE 31

Sodium chloride was formed when 0.0345 mol of chlorine
gas reacted completely with sodium according to the
equation below:

2Na (s) + Cl2 (g) 2NaCl (s)

Calculate the percentage yield if 3.75 g of sodium

chloride is produced in this experiment.

( Ans : 92.90 %)

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1 mol Cl2 Ξ 2 mol NaCl

0.0345 mol Cl2 Ξ 0.0345 mol × 2 mol NaCl
1 mol

= 0.069 mol NaCl

Theoretical mass of NaCl = 0.069 mol x 58.5 gmol-1
= 4.0365 g

Percentage yield = actual yield × 100 %
theoretical yield

= 3.75 g × 100 %
4.0365 g

= 92.90 %

122