Tutorial | MeKa | KumBe | Chapter 1 | Matter

SESSION 2020/2021 TOPIC 1: MATTER
TUTORIAL 1 CHEMISTRY SK015

1.1: Atoms and Molecule

1. a) Definition
i. Number of protons in the nucleus of an atom of an element.
ii. Total number of protons and neutrons in the nucleus of an atom of an
element.
iii. Two or more atoms of the same element having the same number of protons
but different number of neutrons.

b) isotope notation

11P − , 21Q+ , 19 R − 24S
9
,

c) number of sub-atomic particles

Element protons Number of Electrons
14 neutrons 14
silicon 17 17
chlorine 14
20

2. a) Define relative atomic mass.
The mass of an atom in comparison to one twelfth of the mass of one carbon-12 atom
Relative atomic mass = mass of an atom of an element (amu)
1 ×mass of one 12C atom (amu)
12

b) relative atomic mass of atom X

mass of one 12 C atom = 12.00 u

mass of one X atom = 2 (12.00 u) = 24.00 u

Ar of X = 24.00 u = 24.00
1 ×12.00
12 u

c) mass of atom Y in gram

Ar of Y = 32.00
Mass of 1 mol Y = 32.00 g

1 mol Y contains 6.02×1023 Y atoms
6.02×1023 Y atoms = 32.00 g

∴ 1 atom Y = 32.00 g = 5.32×10-23 g
6.023×1023 Y atoms

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015

3. a) average atomic mass of chlorine.

Ratio: 35Cl : 37Cl = 3.127 : 1

3.127 (34.9686) +1(36.9659)

Average atomic mass of Cl =

3.127 +1

= 35.4526 amu

b) average atomic mass of Zr

52(90) + 9(91) +12(92) +14(93) +13(94)

Average atomic mass of Zr =

52 + 9 +12 +14 +13

= 91 amu

1.2: Mole Concept

1. Given: 10.0 g CuSO4.5H2O
a) number of moles of CuSO4.5H2O.

10.0 g
n CuSO4.5H2O = 249.7 g mol−1 = 0.0400 mol

b) number of moles of copper atoms.
1 mol CuSO4.5H2O ≡ 1 mol Cu atom
0.0400 mol CuSO4.5H2O ≡ 0.0400 mol Cu atom

c) mass of anhydrous copper(II) sulphate.
1 mol CuSO4.5H2O ≡ 1 mol CuSO4
0.0400 mol CuSO4.5H2O ≡ 0.0400 mol CuSO4

∴ mass of C atoms = 0.0400 mol x 159.7 g mol-1 = 6.39 g

2. a) Definition
Empirical formula:
formula that shows the simplest ratio of atoms of different elements present in a
molecule

Molecular formula:
formula that shows the actual number of atoms of different elements present in a
molecule

b) Empirical formula C H
(Element

Mass in 100 g 85.7 g 14.3 g
Number of mole
85.7 = 7.142 14.3 = 14.3
Ratio 12.0 =1 1.0 =2

7.142 14.3
7.142 7.142

Empirical formula: CH2

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015
c) Given:
mass of compound = 0.25 g H
volume at STP = 100 mL 0.050
Molar volume at STP = 22.4 L mol-1 0.05 =0.050
1.0
22.4 L CxHy ≡ 1 mol CxHy 0.050 =2
0.1 L ≡ 4.464×10−3 mol CxHy 0.0250

4.464×10−3 mol CxHy ≡ 0.25 g
1 mol CxHy ≡ 56 g

∴ Molar mass of CxHy = 56 gmol-1

n[(12.0 + 2(1.0)] = 56
n=4

∴ Molecular formula: C4H8

3. Given:
mass of CO2 = 1.10 g
mass of H2O = 0.45 g.
molar mass of the hydrocarbon = 84.0 g mol–1.

( )2 1.0 gmol-1
Mass of H in H2O = ×0.45 g = 0.050 g
18.0 gmol-1

12.0 gmol-1
Mass of C in CO2 = ×1.10g = 0.300 g
44.0 gmol-1

Element C
Mass (g)
No of mole(mol) 0.300
0.300 =0.0250
Mol Ratio 12.0

0.0250 =1
0.0250

∴Empirical formula: CH2

n[(12.0 + 2(1.0)] = 84.0
n= 6

∴Molecular formula = C6H12

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CHEMISTRY SK015

4. a) Definition
number of moles of solute per kilogram of solvent.

b) Given
Density of NaOHsolution= Massof solution =1.33 g mL-1
Volumeof solution
Molality of NaOHsolution= No.of moleof solution =10.5 molal
Massof solvent (kg)

i. Mole fraction
Assume mass of solvent (H2O) = 1 kg
Therefore,
n NaOH = 10.5 mol

n H2O = 1000 g = 55.56 mol
18.0 g mol-1

∴ XNaOH = nNaOH

nNaOH + nH2O

= 10.5 mol
66.06 mol

= 0.159

ii. % by mass,
Mass of solute (NaOH) = 10.5 mol × 40.0 g/mol = 420 g
Mass of solvent (H2O) = 1000 g
Mass of solution = 420 g + 1000 g

∴ % by mass = Massof NaOH x100%
Massof solution

= 420 g x100%
1420 g

= 29.6%

ii. Molarity
1420 g

V solution = 1.33 g mL-1 = 1067.7 mL

∴ Molarity = No.of moleof solute
Volumeof solution(L)

10.5 mol
= 1067.7 x10-3 L
= 9.83 mol L−1

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015

c) Given:

Density of H2SO4 solution = Massof solution = 1.84 g mL−1
Volumeof solution

% by mass = Massof solute x100% = 95%
Massof solution

i. Assume mass of solution = 100 g
Therefore, mass of solute(H2SO4) = 95.0 g

n H2SO4 = 95.0 g = 0.968 mol
98.1 g mol-1

V solution = 100 g = 54.3 mL
1.84 g mL-1

Molarity = No.of moleof solute
Volume of solution(L)

0.968 mol
= 54.3 x10-3 L
= 17.8 mol L−1

ii. M1V1 = M2V2
(17.8 M) V1 = (0.50 M)(1.0 L)
V1 = 0.028 L

5. a) Given:
Volume of perfume = 400 mL
Volume of alcohol = 50.0 mL

% by volume = volumeof solute x100%
volumeof solution

= 50.0 mL x100%
400.0 mL

= 12.5%

b) Volume of ether

% by volume = volumeof solute x100%
volumeof solution

40% = volumeof solute (ether ) x100%

30 dm3

V ether = 12 dm3

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SESSION 2020/2021 TOPIC 1: MATTER
1.3: Stoichiometry CHEMISTRY SK015

1. Balance the equations
a) C4H10(g) + O2(g) ⎯→ CO2(g) + H2O(l)
2C4H10(g) + 13O2(g) ⎯→ 8CO2(g) + 10H2O(l)

a) Fe(s) + H2O(l) ⎯→ Fe3O4(s) + H2(g)
3Fe(s) + 4H2O(l) ⎯→ Fe3O4(s) + 4H2(g)

b) Cr(OH)3(aq) + IO3−(aq) ⎯→ CrO32−(aq) + I−(aq) (acidic)

6Cr(OH)3(aq) + IO3−(aq) ⎯→ 6CrO32−(aq) + I−(aq) +12H+(aq) + 3H2O(l)

c) Cl2(aq) ⎯→ ClO4−(aq) + Cl−(aq) (basic)

4Cl2(aq) + 8OH-(aq) ⎯→ ClO4−(aq) + 7Cl−(aq) + 4H2O(l)

2. Given:
mass impure sample of zinc = 50.0 g
V hydrogen gas at room condition = 18.0 L

a) Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2(g)

b) At room condition,
24 L ≡ 1 mol gas
18.0 L ≡ 0.750 mol gas

c) From balance equation,
1 mol H2 gas ≡ 1 mol Zn
0.750 mol H2 gas ≡ 0.750 mol Zn

Mass of Zn = 0.750 mol x 65.4 g/mol
= 49.05 g

% of Zn = Massof Zn x100%
Massof sample

= 49.05 g x100%
50.0 g

= 98.1%

3. Given:
mass of iron ore = 1.55 g
V KMnO4 solution = 92.95 mL
Concentration of KMnO4 solution = 0.0200 M

a) 5Fe2+(aq) + MnO4−(aq) + 8H+(aq) ⎯→ 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

b) percentage of iron in the sample.
Mol of KMnO4 = (0.0200 M)(0.09295 L)
= 1.859 x 10-3 mol

1 mol KMnO4 ≡ 1 mol MnO4-
1.859 x 10-3 mol KMnO4 ≡ 1.859 x 10-3 mol MnO4-

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015
1 mol MnO4- ≡ 5 mol Fe2+
1.859 x 10-3 mol MnO4- ≡ 9.295 x 10-3 mol Fe2+

Mass of iron = 9.295 x 10-3 mol x 55.9 g/mol
= 0.51896 g

% of Fe = Massof iron x100%
Massof iron ore

= 0.5196 g x100%
1.55 g

= 33.5%

4. Given:
Mass of copper(II) oxide, CuO = 7.0 g
V nitric acid, HNO3 solution = 50 mL
Concentration of nitric acid, HNO3 solution = 0.20 M

a) a reactant that is completely consumed during a reaction and limits the amount
of product(s) formed.

b) CuO(s) + 2HNO3(aq) ⎯→ Cu(NO3)2(aq) + H2O(l)

c) Given,
mol of CuO = 0.0881 mol
mol of HNO3 = 0.01 mol

1 mol CuO ≡ 2 mol HNO3
0.0881 mol CuO ≡ 0.1762 mol HNO3
∴ mol of HNO3 given < needed. HNO3 is limiting reactant.

d) expected mass of copper (II) nitrate formed.
2 mol HNO3 ≡ 1 mol Cu(NO3)2
0.01 mol HNO3 ≡ 5 x 10-3 mol Cu(NO3)2

Mass of Cu(NO3)2 = 0.94 g

e) Given:
actual mass of copper(II) nitrate = 0.85 g.

% yield = actual mass x100%
theoritical mass

= 0.85 g x100%
0.94 g

= 90%

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SESSION 2020/2021 TOPIC 1: MATTER
MEKA 1 CHEMISTRY SK015

1.1 Atoms and Molecule

1. a) List of isotopes

24 Mg, 25 Mg, 26 Mg
12 12 12

b) most abundant isotope
24Mg

c) Explanation
They have the same number of protons and electrons but with different number of
neutrons.

d) Value of x

Average atomic mass =  Qimi
 Qi

78.99(24) + x (25) +11.01(26)

24.32 =

78.99 + x +11.01

x = 9.97

2. a) mass spectrum

60.10%

% abundance 39.90%
Abundance

69 71 m/e

b) average atomic mass

(60.10 x 68.93) + (39.90 x70.92)

Average atomic mass =

60.10 + 39.90

= 69.72 amu

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SESSION 2020/2021 TOPIC 1: MATTER
1.2 Mole Concept CHEMISTRY SK015

1. a) mass
Mass MnSO4 = 0.64 mol×151.0 g mol−1

= 97 g

15.8 g
b) n of Fe(ClO4)3 = 354.4 g mol−1

= 0.0446 mol

92.6 g
c) n NH4NO2 = 64.0 g mol−1

= 1.45 mol

1 mol NH4NO2 ≡ 2 mol N
1.45 mol NH4NO2 ≡ 2.9 mol N

1 mol N ≡ 6.02×1023 N atoms
2.9 mol N ≡ 1.75×1024 N atoms

2. Empirical formula Cl O
a) 0.063 0.220
Element
No.of mole 0.063 =1 0.220 = 3.5
Ratio 0.063 0.063

Simplest ratio 1x2=2 3.5x2=7
Empirical Formula
Cl2O7
(b)
Element Si Cl
Mass (g) 2.45 12.4
No.of mole
2.45 = 0.087 mol 12.4 = 0.349 mol
Ratio 28.1 35.5

Empirical Formula 0.087 mol =1 0.349 mol  4
0.087 mol 0.087 mol
3. a) Empirical formula
Assume total mass = 100 g SiCl4
Element
Mass (g) N O
No.of mole 69.55
30.45
Ratio 69.55 = 4.35 mol
30.45 = 2.175 mol 16.0
Empirical Formula 14.0
4.35 = 2
2.175 =1 2.175
2.175

NO2

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b) molecular formula

n[NO2 ]= 90
n = 90 =1.96

46
2

∴ Molecular formula = N2O4

4. a) mass of solute

Molarity = No.of mole of solute (Ca(CH3COO−)2)
Volume of solution (L)

0.267 mol L-1 = No.of mole of solute (Ca(CH3COO− )2) 0.0496 mol
185.810−3 L

n of solute = 0.0496 mol

Mass of solute = 0.0496 mol × 158.1 g mol-1
= 7.84 g

b) Molarity

21.1g
n of potassium iodide = 166 g mol−1 = 0.127 mol

No.of mole of solute

Molarity = Volume of solution (L)

0.127 mol
= 500x10−3 L

= 0.254 M

c) mol of solute

No.of mole of solute

molarity = Volume of solution (L)

No.of mole of solute

0.850 M =

145.6 L

n solute = 124 mol

e) M1V1 = M2V2
(0.25 M)(37.00 mL) = M2 (150 mL)
M2 = 0.06167 M

f) M1V1 = M2V2
(2.86×10-3 M)(350 mL) = (1.63 M)V2
V2 = 0.614 mL

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SESSION 2020/2021 TOPIC 1: MATTER
1.3 Stoichiometry CHEMISTRY SK015

1. a) 16Cu(s) + S8(s) →8Cu2S(s)
b) H3PO4(aq) + 2NaOH(aq) → Na2HPO4(aq) + 2H2O(l)
c) 2Co3+(aq) + 2H20(l) + Mn2+(aq) → 2Co2+(aq) + MnO2(s) + 4H+(aq)
d) 2MnO4-(aq) + H2O(l) + 3ClO3-(aq) → 2MnO2(s) + 3ClO4-(aq) + OH-(aq)

2. Mass of each product form
Balance the equation:
B2H6(g) + 6H2O(l)→2H3BO3(s) + 6H2(g)

33.61g
n B2H6 = 27.6 g mol−1 = 1.218 mol

From balanced equation,
1 mol B2H6 ≡ 2 mol H3BO3
1.218 mol B2H6 ≡ 2.436 mol H3BO3

Mass of H3BO3 = (2.436 mol)(61.8 g mol-1)= 150.5 g

From balanced equation,
1 mol B2H6 ≡ 6 mol H2
1.218 mol B2H6 ≡ 7.308 mol H2

Mass of H2 = (7.308 mol)(2.0 g mol-1)= 14.62 g

3. a) mass of Fe3O4
Given,

24.5 g
n Fe = 55.9 g mol-1 = 0.438 mol

12.6 g
n H2O = 18.0 g mol−1 =0.700 mol

From balance equation,
3 mol Fe ≡ 4 mol H2O
0.438 mol Fe ≡ 0.584 mol H2O needed

n H2O needed < n H2O given
∴ Fe is limiting reactant

From balanced equation,
3 mol Fe ≡ 1 mol Fe3O4
0.438 mol Fe ≡ 0.146 mol Fe3O4

Mass of Fe3O4 = (0.146 mol)(231.55 g mol-1) = 33.81 g

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015
b) volume of H2
From balanced equation,
3 mol Fe ≡ 1 mol H2
0.438 mol Fe ≡ 0.584 mol H2

1 mol H2 ≡ 22.4 L
0.584 mol H2 ≡ 13.1 L

5. Given;

2.80 g
n O2 = 2(16.0) g mol−1 = 0.0875 mol

4.20 g
n Ca = 40.1g mol−1 = 0.105 mol

a) mole of CaO
From balanced equation
2 mol Ca ≡ 2 mol CaO
0.105 mol Ca ≡ 0.105 mol CaO

b) mole of CaO
1 mol O2 ≡ 2 mol CaO
0.0875 mol O2 ≡ 0.175 mol CaO

c) By comparing 4(a) and 4(b), the smallest mol of CaO is produced from Ca,
therefore, Ca is a limiting reactant.

d) mass of CaO
mass CaO = (0.105 mol) (56.1 g mol-1)
= 5.89 g

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SESSION 2020/2021 TOPIC 1: MATTER
KUMBE 1 CHEMISTRY SK015

1.1 Atoms and Molecules

1. a) definition
Relative atomic mass,Ar is defined as the mass of one atom of an element compared
to 1/2 the mass of one atom of 12C with the mass 12.00 amu.

b) most abundance isotope
20Ne

c) Number of protons = 10
Number of neutron= 10

2. a) 13.003

= 1.084 times

12.000

b) mass of one 12C
1 mol C ≡ 6.02×1023 atoms C
1 mol C ≡ 12 g
6.02×1023 atoms C≡ 12 g
1 atom C ≡ 1.993×10-23 g

c) % of 12C
12.011 = 12.000 a + 13.003 (1-a)
a = 0.989
= 98.9%

1.2 Mole Concept

1. a) Molecular mass = 20(12) + 24(1) + 2(14) + 2(16) = 324 amu

b) no.of moles = 1.08g =3.33×10-3 mol
324g mol-1

c) molecules of quinine
1 mol C20H24N2O2≡6.02×1023 molecules
3.33×10-3 mol C20H24N2O2≡2.01×1021 molecules

d) number of H atoms
1 mol C20H24N2O2≡24 mol H
3.33×10-3 mol C20H24N2O2 ≡ 0.0799 mol H

1 mol H ≡ 6.02×1023 atoms
0.0799 mol H ≡ 4.82×1022 atoms

e) mass of C
1 mol C20H24N2O2≡20 mol C
3.33×10-3 mol C20H24N2O2 ≡ 0.06666 mol C

Mass of C = (0.06666 mol)(12 g mol-1)
= 0.800 g

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015
2. value of x

7.95g

n K2CO3 = 2(39.1) +12+3(16) = 0.05753 mol

n K2CO3.xH2O ≡ n K2CO3 = 0.05753 mol

10

2(39.1) +12+3(16) +x (2(1) +16) =0.05753 mol

x=2

3. molecular formula

12
mass of C in CO2 = x 0.361g =0.0985 g

44

2
mass of H in H2O = x 0.147g = 0.0163 g

18

mass of O = 0.202 g – 0.0985 g – 0.0163 g = 0.0872 g

C H O
0.0163
Mass (g) 0.0985 0.0872
0.0163
Moles 0.0163 = 8.204 × 10−3 1 = 0.0163 0.0872 = 5.451 × 10−3
12 16
Simplest 2.99
ratio 1.505 1
6
3 2

The empirical formula : C3H6O2
The molecular formula :

n (3(12) +6(1) +2(16)) =148

n= 2

The molecular formula : C6H12O4

4. Given,

Density of solution = mass of solution = 1.00 g mL-1

Volume of solution

%by mass = mass of solute ×100% = 2.5%

mass of solution

a) concentration
Assume mass of solution = 100 g

V solution = 100 mL
Mass of solute = 2.5 g

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CHEMISTRY SK015

Mass of solvent = mass of solution – mass of solute
= 100 g – 2.5 g

= 97.5 g

n of solute (ethanol) = 2.5g = 0.0543 mol

46 gmol-1

no.of moles of solute

molarity = volume of solution (L)

0.0543 mol

=

0.100 L

= 0.543 mol L-1

b) mass of ethanol
100 mL beer contain 2.5 g ethanol
200 mL beer contain 5.00 g ethanol

5. Molar mass
Given

no.of moles of solute

Molarity = volume of solution (L) = 6.5 M

Density = mass of solution = 0.888 g cm-3

Volume of solution

Molar mass = 98 g mol-1

Assume V solution = 1 L
n solute(HA) = 6.5 mol
mass solute = (6.5 mol)(98 g mol-1) = 637 g

mass solution = 888 g
mass solvent = 888 g – 637 g

= 251 g

n solute

Molality =

mass of solution (kg)
6.5 mol

=

0.251 kg

= 26 mol kg-1

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CHEMISTRY SK015

6. Given;

Density solution = mass of solution = 1.23 g mL-1

volume of solution

% by mass = mass of solute x100% = 90%
mass of solution

Assume mass solution = 100 g
Therefore,
Mass solute = 90.0 g
V solution = 81.3 mL

a) molarity

90.0 g
n CH3COOH = 60 g mol-1 = 1.50 mol

no.of moles of solute

molarity = volume of solution (L)

1.50 mol

=

0.0813 L

= 18.5 M

b) volume

n dilute CH3COOH = (4.00 M)(1.8 L) = 7.2 mol

7.2 mol
V concentrated CH3COOH = 18.5 mol L-1

= 0.39 L

7. Given

Density solution = mass of solution = 0.182 g cm-3

volume of solution

%by mass = mass of solute x100% = 96.0%
mass of solution

a) molarity
Assume mass solution = 100 g
Mass solute = 96.0 g
V solution = 549.5 cm3
n solute = 0.980 mol

no.of moles of solute

molarity =

volume of solution (dm3)
0.980 mol

= 0.5495 dm3

= 1.78 M

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015
b) Volume required
2SO4( ) + Zn(s) → 2 ( ) + ZnSO4(

1 mol H2 ≡ 24 dm3
8.396×10-3 mol H2 ≡ 0.2015 dm3

From eq,
1 mol H2 ≡ 1 mol H2SO4
8.396×10-3 mol H2 ≡ 8.396×10-3 mol H2SO4

no.of moles of solute

Molarity =

volume of solution (dm3)
8.39610−3 mol

1.78 M = volume of solution (dm3)

V solution = 4.72×10-3 dm3

8. a) Step to dilute stock solution
1. Fill approximately 50 mL of distilled water in to a 100 mL beaker.
2. Rinse a 10 mL pipette with HCl stock solution.
3. Pipette 5 mL of the stock solution and transfer it into the beaker containing 50 mL
distilled water.
4. Stir the solution with a glass rod and transfer it into a 250 mL volumetric flask.
5. Add distilled water to the volumetric flask up to the graduated mark. Stopper and
shake the flask to obtain homogeneous solution.

b) Volume diluted muriatic acid

0.15 g
n of HCl (15 g) = 36.5 g mol-1 = 4.110×10-3 mol

no.of moles of solute
molarity = volume of solution (dm3)

4.1110−3mol
0.2 M = volume of solution (dm3)

V solution = 0.0206 L = 20.6 mL

1.3 Stoichiometry

1. balanced equation

Oxidation:(C2O42- → 2CO2 + 2e)x5
Reduction:(MnO4- +8H+ + 5e → Mn2+ + 4H2O)x2
Overall reaction: 5C2O42- + 2MnO42- +16H+ → 10CO2 + 2Mn2+ +8H2O

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015
2. a) ionic equation
For step 3:

Oxidation:2S2O32- → S4O62- +2e
Reduction:I3- + 2e → 3I-
Overall reaction: 2S2O32- + I3- → S4O62- + 3I-

For step 2:

2CuSO4 + 5I- → 2CuI + I3- + 2SO42-

b) % of Cu in bronze
n S2O32- = (1.05 M)(0.0262 L) = 0.02751 mol

From equation,
2 mol S2O32- ≡ 1 mol I3-
0.02751 mol S2O32- ≡ 0.01376 mol I3-

1 mol I3- ≡ 2 mol CuSO4
0.01376 mol I3- ≡ 0.02751 mol CuSO4
Mass of Cu = (0.02751 mol)(63.6 g mol-1) = 1.7496 g

% of Cu = 1.7496 g x100 = 46%
3.8 g

3. a) mass of ammonia
Given:

224 g
n NH4Cl = 53.5g mol−1 = 4.187 mol

112 g
n CaO = 56.1g mol-1 = 1.996 mol

Needed,
From balance equation,
2 mol NH4Cl ≡ 1 mol CaO
4.187 mol NH4Cl ≡ 2.094 mol CaO

n CaO given < n CaO needed
CaO is limiting reactant

From balance equation,
1 mol CaO ≡ 2 mol NH3
1.996 mol CaO ≡ 3.992 mol NH3

Mass of NH3 = (3.992 mol)(17 g mol-1)
= 67.9 g

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CHEMISTRY SK015
b) mass of excess reactant remain
1 mol CaO ≡ 2 mol NH4Cl
1.996 mol CaO ≡ 3.992 mol NH4Cl

Mass of NH4Cl reacted = (3.992 mol)(53.5 g mol-1) = 213.6 g
Mass of NH4Cl remained = 224.0 g – 213.6 g = 10.4 g

4. a) i. balanced equation

MO2 (s) + 4HCl(aq) → MCl2 (aq) + Cl2(g) + 2H2O(l)

ii. At STP,
22.4 L ≡ 1 mol Cl2
0.386 L ≡ 0.0172 mol Cl2

From balanced equation,
1 mol Cl2 ≡ 1 mol MO2
0.0172 mol Cl2 ≡ 0.0172 mol MO2

Molar mass MO2 = 1.5 g = 87 g mol-1

0.0172 mol

∴ relative molecular mass = 87
∴ relative atomic mass = 87-(2×16) = 55

b) i. limiting reactant
Given;

0.20 g
n MO2 = 8.7 x101 = 2.30×10-3 mol

n HCl = ( 0.10 ) ( 25) = 2.50×10-3 mol

1000

needed,

from balanced equation
1 mol MO2 ≡ 4 mol HCl
2.30×10-3 mol MO2 ≡ 9.20×10-3 mol HCl

n HCl given < n HCl needed
∴ HCl is limiting reactant

ii. mass of MCl2 produce
4 mol HCl ≡ 1 mol MCl2
2.50×10-3 mol HCl ≡ 6.30×10-3 mol MCl2

Mass of MCl2 = (6.30×10-3 mol)(126 g mol-1)
= 0.079 g

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015
iii. Percentage yield

% yield = actual yield x100 %
theoritical yield

= 0.072 g x100 %
0.079 g

= 91%

5. % yield

25.0 g
n C6H12 = 6(12) +12(1) g mol−1 = 0.2976 mol

From balance equation,
2 mol C6H12 ≡ 2 mol H2C6H8O4
0.2976 mol C6H12 ≡ 0.2976 mol H2C6H8O4

Mass of H2C6H8O4 = (0.2976 mol)(146 g mol-1)
= 43.45 g

% yield = actual yield x100 %
theoritical yield

= 33.5g x100 %
43.45 g

= 77.1%

6. a) Oxidation :(2I− → I2 + 2e ) x 5

_R_e_d_u_c_t_i_o_n_:_(_2_IO__3_−_+_1_2_H__+_+__1_0_e_→___I_2 _+_6_H__2_O_)_____________

Overall reaction :2IO3− +10I− +12H+ →6I2 + 6H2O

IO3− + 5I− + 6H+ →3I2 + 3H2O

b) mass of KIO3
n S2O32- = (0.2 M)(0.0537 L) = 0.0107 mol

From balanced equation,
2 mol S2O32- ≡ 1 mol I2
0.0107 mol S2O32- ≡ 5.35×10-3 mol I2

From balanced equation,
3 mol I2≡ 1 mol IO3-
5.35×10-3 mol I2 ≡ 1.78×10-3 mol IO3-

n KIO3 ≡ n IO3- ≡ 1.78×10-3 mol

mass of KIO3 = (1.78×10-3 mol)(214 g mol-1)
= 0.38 g

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SESSION 2020/2021 TOPIC 1: MATTER
CHEMISTRY SK015
7. a) mass of NH3
Given:

33.0 g
n NH4Cl = 53.5 g mol−1 = 0.617 mol

33.0 g
n Ca(OH)2 = 74.1g mol−1 = 0.445 mol

From balance equation,
2 mol NH4Cl ≡ 1 mol Ca(OH)2
0.617 mol NH4Cl ≡ 0.309 mol Ca(OH)2 (needed)

n Ca(OH)2 needed < n Ca(OH)2 given
Ca(OH)2 is the excess reactant while NH4Cl is limiting reactant.

From balance equation,
2 mol NH4Cl ≡ 2 mol NH3
0.617 mol NH4Cl ≡ 0.617 mol NH3

Mass of NH3 = (0.617 mol)(17 g mol-1)
= 10.5 g

b) mass of excess reactant remained
n Ca(OH)2 used = 0.309 mol
n Ca(OH)2 excess = (0.445-0.309) mol = 0.136 mol
mass of Ca(OH)2 excess = (0.136 mol)(74.1 g mol-1) = 10.1 g

8. mass of Na2CS3
Mass of CS2 = (1.26 g mL-1)(92.5 mL) = 116.6 g

116.6 g
n CS2 = 76.2 g mol-1 = 1.53 mol

From balance equation,
6 mol NaOH ≡ 3 mol CS2
2.78 mol NaOH ≡ 1.39 mol CS2 (needed)

n < nCS2 (needed) CS2 (provided)

CS2 is the excess reactant

 NaOH is the limiting reactant

6 mol NaOH  2 mol Na2CS2
2.78 mol NaOH  0.9267 mol Na2CS2
Mass of Na2CS2 = 0.9267 mol x154.3g mol-1 =143g

6 mol NaOH  2 mol Na2CS2
2.78 mol NaOH  0.9267 mol Na2CS2

Mass of Na2CS3 = (0.9267 mol)(154.3 g mol-1) = 143 g

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