BAB
3
3.1
Konsep Mol, Formula dan Persamaan Kimia
The Mole Concept, Chemical Formulae and
Equations
Modul
PBD
Jisim Atom Relatif dan Jisim Molekul Relatif
Relative Atomic Mass and Relative Molecular Mass
1 Hitung jisim formula relatif setiap bahan berikut.
Determine the relative formula mass of each of the following substance.
SP 3.1.2 TP 2
[Jisim atom relatif/Relative atomic mass: H = 1, N = 14, O = 16, Na = 23, Mg = 24, P = 31, S = 32,
Cu = 64]
Bahan
Substance
Jisim formula relatif
Relative formula mass
(a) Ammonium fosfat, (NH4)3PO4
Ammonium phosphate, (NH4)3PO4
3N + 12H + P + 4O = 3(14) + 12(1) + 31 + 4(16)
= 149
(b) Natrium karbonat kontang, Na2CO3
Anhydrous sodium carbonate, Na2CO3
2Na + C + 3O = 2(23) + 12 + 3(16)
= 106
(c) Natrium karbonat terhidrat, Na2CO3.10H2O
Hydrated sodium carbonate, Na2CO3.10H2O
2Na + C + 3O + 10(H2O) = 2(23) + 12 + 3(16) +
10(18) = 286
(d) Kuprum sulfat kontang, CuSO4
Anhydrous copper sulphate, CuSO4
Cu + S + 4O = 64 + 32 + 4(16) = 160
(e) Kuprum sulfat terhidrat, CuSO4.5H2O
Hydrated copper sulphate, CuSO4.5H2O
Cu + S + 4O + 5(18) = 64 + 32 + 4(16) + 90
= 250
3.2
Konsep Mol/Mole Concept
2 Tentukan bilangan zarah dalam bahan-bahan berikut.
Determine the number of particles in the following substances.
SP 3.2.2 TP 2
(a)
0.5 mol kalsium, Ca
0.5 mol calcium, Ca
0.5 x 6.02 x 1023 = 3.01 x 1023 atom kalsium/calcium atoms
(b)
1.0 mol gas oksigen, O2
1.0 mol oxygen gas, O2
1.0 x 6.02 x 1023 = 1.20 x 1024 molekul oksigen/oxygen molecules
2 x 1.0 x 6.02 x 1023 = 6.02 x 1023 atom oksigen/oxygen atoms
(c)
2.0 mol air, H2O
2.0 mol water, H2O
2.0 x 6.02 x 1023 = 1.20 x 1024 molekul air/water molecules
2 x 2.0 x 6.02 x 1023 = 2.40 x 1024 atom hidrogen/hydrogen atoms
1 x 2.0 x 6.02 x 1023 = 1.20 x 1024 atom oksigen/oxygen atoms
(d)
3.0 mol aluminium
klorida, AlCl3
3.0 mol aluminium
chloride, AlCl3
3.0 x 6.02 x 1023 = 1.80 x 1024 unit formula/formula units
1 x 3.0 x 6.02 x 1023 = 1.81 x 1024 ion aluminium/aluminium ions
3 x 3.0 x 6.02 x 1023 = 5.42 x 1024 ion klorida/chloride ions
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3 Nyatakan bilangan molekul unsur berikut berdasarkan bilangan mol yang diberikan.
State the number of molecules of the following elements based on the number of moles given.
SP 3.2.2 TP 2
(a)
0.5 mol kalsium, Ca
0.5 mol calcium, Ca
0.5 x 6.02 x 1023 = 3.01 x 1023 atom kalsium/calcium atoms
(b)
1.0 mol gas oksigen, O2
1.0 mol oxygen gas, O2
1.0 x 6.02 x 1023 = 6.02 x 1023 molekul oksigen/oxygen molecules
2 x 1.0 x 6.02 x 1023 = 1.20 x 1024 atom oksigen/oxygen atoms
(c)
2.0 mol air, H2O
2.0 mol water, H2O
2.0 x 6.02 x 1023 = 1.20 x 1024 molekul air/water molecules
2 x 2.0 x 6.02 x 1023 = 2.40 x 1024 atom hidrogen/hydrogen atoms
1 x 2.0 x 6.02 x 1023 = 1.20 x 1024 atom oksigen/oxygen atoms
(d)
3.0 mol aluminium
klorida, AlCl3
3.0 mol aluminium
chloride, AlCl3
3.0 x 6.02 x 1023 = 1.81 x 1024 unit formula/formula units
1 x 3.0 x 6.02 x 1023 = 1.81 x 1024 ion aluminium/aluminium ions
3 x 3.0 x 6.02 x 1023 = 5.42 x 1024 ion klorida/chloride ions
4 Hitung jisim setiap bahan berikut.
Calculate the mass of each substance below.
SP 3.2.4 TP 2
[Jisim atom relatif/ Atomic relative mass: H = 1, C = 12, O = 16, S = 32, Ca = 40, Br = 80, Pb = 207]
0.2 mol plumbum, Pb
0.2 mol lead, Pb
0.2 x 207 g mol–1 = 41.4 g
0.8 mol karbon dioksida, CO2
0.8 mol carbon dioxide, CO2
0.8 x [12 + 2(16)] g mol–1 = 35.2 g
1.5 mol kalsium bromida, CaBr2
1.5 mol calcium bromide, CaBr2
1.5 x [40 + 2(80)] g mol–1 = 300.0 g
3.0 mol asid sulfurik, H2SO4
3.0 mol sulphuric acid, H2SO4
3.0 x [2(1) + 32 + 4(16)] g mol–1 = 294.0 g
5 Hitung isi padu setiap gas berikut.
Calculate the volume of each gas below.
SP 3.2.5 TP 2
0.05 mol gas klorin, Cl2 pada STP
0.05 mol chlorine gas, Cl2 at STP
0.05 x 22.4 dm3 mol–1 = 1.12 dm3
0.60 mol gas helium, He pada STP
0.60 mol helium gas, He at STP
0.60 x 22.4 dm3 mol–1 = 13.44 dm3
1.75 mol gas nitrogen, N2 pada keadaan bilik
1.75 mol nitrogen gas, N2 at room conditions
1.75 x 24.0 dm3 mol–1 = 42.0 dm3
4.0 mol gas sulphur dioksida, SO2 pada keadaan
bilik
4.0 mol sulphur dioxide gas, SO2 at room conditions
4.0 x 24.0 dm3 mol–1 = 96.0 dm3
6 Beberapa biji belon diisi gas berlainan pada keadaan bilik. Lengkapkan jadual berikut.
Several balloons are filled with different gases at room conditions. Complete the following table.
SP 3.2.7 TP 4
[Jisim atom relatif/Relative atomic mass: C = 12, O = 16, Ar = 40; Pemalar Avogadro/Avogadro Constant =
6.02 x 1023; Isi padu molar gas pada keadaan bilik/Molar volume of gas at room conditions = 24 000 cm3]
Menganalisis
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Belon
Balloon
0.8 g
O2
8.0
g gg
2.2
0.8
Ar
CO
O22
2.2
0.8gg
CO
O2
8.0
gg
2.2
ArCO2
8.0 g
Ar
Bilangan mol
Number of moles
= 0.025
= 0.05
= 0.20
Bilangan molekul
Number of molecules
0.025 x 6.02 x 1023
= 1.51 x 1022
0.05 x 6.02 x 1023
= 3.01 x 1022
0.20 x 6.02 x 1023
= 1.20 x 1023
Isi padu gas
Volume of gas
0.025 x 24 000
= 600 cm3
0.05 x 24 000
= 1200 cm3
0.20 x 24 000
= 4800 cm3
Jika jisim 18 dm3 XO2 pada keadaan bilik adalah 34.5 g, hitung jisim satu mol XO2. Kemudian, hitung
jisim atom relatif X dan kenal pasti unsur X.
If the mass of 18 dm3 XO2 at room conditions is 34.5 g, calculate the mass of one mole of XO2. Then, calculate
the relative atomic mass of X and identify element X.
Bilangan mol XO2
Number of moles of XO2
= = 0.75
Jisim 1 mol XO2
= = 46.0
Mass of 1 mol XO2
X + 2O = 46.0
X = 46.0 – 32 = 14.0
X = nitrogen
3.3
Formula Kimia/Chemical Formulae
7 Tentukan formula empirik setiap bahan berikut.
Determine the empirical formula of each of the following substances.
Sebatian
Compound
SP 3.3.1
Formula molekul
Molecular Formula
Formula empirik
Empirical Formula
NH3
NH3
C4H8O2
C2H4O
Propena/Propene
C3H6
CH2
Naftalena/Naphthalene
C10H8
C5H4
Aseton/Acetone
C3H6O
C3H6O
Ammonia
Asid butanoik/Butanoic acid
1.28 g X bergabung dengan 0.16 g oksigen untuk membentuk suatu oksida. Jika jisim atom relatif X dan
oksigen adalah 64 dan 16 masing-masing, apakah formula empirik sebatian ini?
1.28 g X combined with 0.16 g oxygen to form an oxide. If the relative atomic mass of X and oxygen are 64 and
16 respectively, what is the empirical formula of the compound? TP 3
8
Unsur/Element
X
O
Jisim/Mass (g)
1.28
0.16
= 0.02
= 0.01
2
1
Bilangan mol
Number of moles
Nisbah mol paling ringkas
Simplest mol ratio
Formula empirik/Empirical formula = X2O
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Formula empirik suatu sebatian ialah M2O3. Jika jisim M dalam suatu sampel sebatian adalah 0.81 g,
berapakah jisim oksigen dalam sampel itu? [Jisim atom relatif: O = 16, M = 27]
The empirical formula of a compound is M2O3. If the mass of M in a sample of the compound is 0.81 g, what is
the mass of oxygen in the sample? [Relative atomic mass: O = 16, M = 27] TP 4
9
Mol M = = 0.030
Mol O = x 0.03 = 0.045
Jisim/Mass O = 0.045 x 16 = 0.72 g
Komposisi sebatian Q mengikut peratusan jisim adalah 38.71% karbn, 9.68% hidrogen dan 51.61%
oksigen. Tentukan formula empirik dan formula molekul Q jika jisim molekul relatifnya adalah 62.
The percentage composition by mass of compound Q is 38.71% carbon, 9.68% hydrogen and 51.61% oxygen.
Determine the empirical and molecular formulas of Q if its relative molecular mass is 62. TP 4
10
Unsur/Element
C
H
O
Jisim dalam 100 g sebatian (g)
Mass in 100 g compound (g)
38.71
9.68
51.61
Bilangan mol
Number of moles
= 3.23
= 9.68
= 3.23
1
3
1
Nisbah mol paling ringkas
Simplest mol ratio
Formula empirik/Empirical formula = CH3O
(CH3O)n = 62
[12 + 3(1) + 16]n = 62
n = = 2
Formula molekul/Molecular formula = C2H6O2
11 Lengkapkan laporan aktiviti berikut.
Complete the following activity report.
Aktiviti 3.1
SP 3.3.2 TP 6
Tujuan: Untuk menentukan formula empirik magnesium oksida
Aim: To determine the empirical formula of magnesium oxide
Pernyataan masalah: Bagaimanakah formula empirik magnesium oksida ditentukan?
Problem statement: How to determine the empirical formula of magnesium oxide?
Bahan dan radas: Kertas pasir, mangkuk pijar dengan penutup, alas segi tiga tanah liat, tungku kaki tiga,
penunu Bunsen, neraca, penyepit, pita magnesium
Materials and apparatus: Sandpaper, crucible with lid, clay pipe triangle, tripod stand, Bunsen burner, weighing
balance, tongs, magnesium ribbon
Prosedur/Procedure:
1
Timbang sebuah mangkuk pijar yang bersih bersama-sama dengan penutupnya dan catatkan
jisimnya.
Weigh a clean crucible together with its lid and record its mass.
2 Bersihkan pita magnesium yang berukuran lebih kurang 20 cm dengan kertas pasir.
Clean a magnesium ribbon that is approximately 20 cm in length with sandpaper.
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3 Gulungkan pita magnesium secara longgar dan letakkannya di dalam mangkuk pijar.
Coil the magnesium ribbon loosely and place it in the crucible.
4 Timbang mangkuk pijar berpenutup yang berisi pita magnesium dan catatkan jisimnya.
Weigh the crucible and lid containing the magnesium ribbon and record its mass.
5 Letakkan mangkuk pijar di atas alas segi tiga tanah liat dan tungku kaki tiga.
Put the crucible on top of a clay pipe triangle and tripod stand.
6
Panaskan mangkuk pijar tanpa penutupnya secara sederhana dan kemudiannya dengan kuat. Sebaik
sahaja pita magnesium mula terbakar, tutup mangkuk pijar dengan penutupnya.
Heat the crucible without its lid on a medium flame and then strongly. When the magnesium ribbon starts to
burn, cover the crucible with its lid
7
Teruskan pemanasan dengan kuat. Buka penutup mangkuk pijar sekali-sekala tetapi pastikan bahawa
asap putih tidak terbebas.
Continue heating strongly. Open the lid at intervals and ensure that no white smoke escaped.
8
Apabila pemanasan magnesium itu kelihatan telah selesai, biarkan mangkuk pijar dan kandungannya
menyejuk ke suhu bilik. Timbang mangkuk pijar dan kandungannya Bersama-sama dengan
penutupnya sekali lagi. Catatkan jisimnya.
When the magnesium is completely burnt, the crucible is allowed to cool to the room temperature. Weigh the
crucible with its lid and its content again. Record the mass.
9
Ulang langkah pemanasan, penyejukan dan penimbangan sehingga memperoleh bacaan jisim hasil
yang tetap.
Repeat the heating, cooling and weighing steps until a constant mass is obtained.
penutup
lid
mangkuk pijar
crucible
pita megnesium
magensium ribbon
Keputusan/Results:
Jisim mangkuk pijar + penutup (g), x1
Mass of crucible + lid (g)
Jisim mangkuk pijar + penutup + magnesium (g), x2
Mass of crucible + lid + magnesium (g)
Jawapan murid
Student’s answers
Jisim mangkuk pijar + penutup + magnesium oksida (g), x3
Mass of crucible + lid + magnesium oxide (g)
Warna magnesium oksida
Colour of magnesium oxide
Putih
White
Pengiraan/Calculations:
(a) Jisim magnesium yang digunakan, y1
Mass of magnesium used
=
x2 – x1
(b) Bilangan mol magnesium yang digunakan, y2
Number of moles of magnesium used
=
y1
24
(c) Jisim oksigen yang berpadu dengan magnesium, y3
Mass of oxygen combined with magnesium
=
x3 – x2
(d) Bilangan mol oksigen yang berpadu dengan magnesium, y4
Number of moles of oxygen combined with magnesium
=
y3
16
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(e) Nisbah mol magnesium kepada mol oksigen dalam magnesium oksida
Mole ratio of magnesium to oxygen in magnesium oxide
=
y2: y4
(f) Formula empirik magnesium oksida
Empirical formula of magnesium oxide
=
MgO Perbincangan/Discussion:
Pita magnesium perlu dibersihkan dengan kertas pasir untuk menghilangkan lapisan oksida .
The magnesium ribbon must be cleaned with sandpaper to remove the layer of oxide .
Mengapakah mangkuk pijar ditutup apabila magnesium mula membakar?
Why was the crucible closed when magnesium starts to burn?
Untuk mengelakkan kehilangan magnesium oksida dalam bentuk asap putih.
To prevent the loss of magnesium oxide in the form of white fumes.
1
2
3 Mengapakah penutup mangkuk pijar dibuka sekali sekala semasa pemanasan magnesium?
Why was the crucible lid opened once in a while during the heating of magnesium?
Untuk membenarkan udara masuk supaya magnesium terbakar dengan lengkap.
To allow air to enter so that magnesium burns completely.
4 Bagaimanakah anda dapat memastikan bahawa pita magnesium telah dibakar dengan lengkap?
How can you ensure that the magnesium ribbon has been completely burned?
Langkah pemanasan , penyejukan dan penimbangan diulang sehingga nilai timbangan
jisim tetap.
The heating , cooling and weighing steps are repeated until a fixed mass value is
obtained.
5
Apakah inferens yang boleh anda buat mengenai nisbah mol magnesium dan oksigen yang berpadu
dengan formula empirik magnesium oksida?
What inference can you make regarding the mole ratio of magnesium and oxygen that combined with the
empirical formula of magnesium oxide?
Nisbah mol magnesium dan oksigen yang berpadu adalah sama dengan formula empirik
magnesium oksida.
The mole ratio of magnesium and oxygen that combined is same as the empirical formula of
magnesium oxide.
6 Tuliskan persamaan kimia bagi tindak balas antara magnesium dengan oksigen dalam eksperimen ini.
Write the chemical equation for the reaction between magnesium and oxygen in this experiment.
2Mg(p/s) + O2(g) → 2MgO(p/s)
7
Seorang pelajar mendapati bahawa formula empirik yang didapatinya bagi magnesium oksida adalah
kurang tepat. Cadangkan dua sebab untuk ketidaktepatan ini
A student found that the empirical formula for magnesium oxide that he obtained is not accurate. Suggest
two reasons for this inaccuracy.
1 Pembakaran magnesium tidak lengkap
Incomplete burning of magnesium
2 Magnesium oksida hilang dalam bentuk wasap putih
Magnesium oxide was lost in the form of white fumes
Kesimpulan/Conclusion:
Formula empirik magnesium oksida ialah MgO
The empirical formula of magnesium oxide is MgO
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3.4
Persamaan Kimia/Chemical Equations
12 Imbangkan setiap persamaan kimia berikut.
Balance each of the following chemical equation.
SP 3.4.1 TP 2
Persamaan kimia tidak seimbang
Unbalanced chemical equation
Persamaan kimia seimbang
Balanced chemical equation
(a) K + H2O → KOH + H2
2K + 2H2O → 2KOH + H2
(b) Ba(OH)2 + H2SO4 → BaSO4 + H2O
Ba(OH)2 + H2SO4 → BaSO4 + 2H2O
(c) CaCO3 + HCl → CaCl2 + CO2 + H2O
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
(d) NH3 + H2SO4 → (NH4)2SO4
2NH3 + H2SO4 → (NH4)2SO4
(e) H3PO4 + ZnO → Zn3(PO4)2 + H2O
2H3PO4 + 3ZnO → Zn3(PO4)2 + 3H2O
(f) NH3 + O2 →NO + H2O
4NH3 + 5O2 → 4NO + 6H2O
(g) NaNO3 → NaNO2 + O2
2NaNO3 → 2NaNO2 + O2
(h) Mg + HNO3 → Mg (NO3)2 + H2
Mg + 2HNO3 → Mg(NO3)2 + H2
(i) PbO + NH3 → Pb + H2O + N2
3PbO + 2NH3 → 3Pb + 3H2O + N2
(i) C6H12O6 + O2 → CO2 + H2O
C6H12O6 + 6O2 →6CO2 + 6H2O
13 Tuliskan persamaan kimia termasuk keadaan fizik untuk setiap tindak balas berikut.
Write a chemical equation including the physical states for each of the following reactions.
SP 3.4.2 TP 2
(a) Asid nitrik cair meneutralkan larutan akues kalium hidroksida untuk membentuk garam kalium
nitrat.
Dilute nitric acid neutralises aqueous potassium hydroxide to form potassium nitrate salt.
KOH(ak/aq) + HNO3(ak/aq) → KNO3(ak/aq) + H2O(ce/l) (b) Apabila larutan akueus kalsium hidroksida ditambah kepada asid sulfurik cair, mendakan putih
kalsium sulfat terhasil.
When aqueous calcium hydroxide is added to dilute sulphuric acid, a white precipitate of calcium sulphate
is obtained.
Ca(OH)2(ak/aq) + H2SO4(ak/aq) → CaSO4(p/s) + 2H2O(ce/l) (c) Apabila serbuk natrium karbonat dilarutkan dalam asid hidroklorik cair, larutan akueus natrium
klorida didapati. Gas karbon dioksida dibebaskan.
When powdered sodium carbonate is dissolved in dilute hydrochloric acid, an aqueous solution of sodium
chloride is obtained. Carbon dioxide gas is liberated.
Na2CO3(p/s) + 2HCl(ak/aq) → 2NaCl(ak/aq) + H2O(ce/l) + CO2(g)
(d) Apabila larutan akueus plumbum nitrat ditambah kepada larutan akueus kalium iodida, suatu
mendakan kuning plumbum iodida terbentuk.
When aqueous lead nitrate is added to aqueous potassium iodide, a yellow precipitate of lead iodide is
formed.
Pb(NO3)2(ak/aq) + 2KI(ak/aq) → PbI2(p/s) + 2KNO3(ak/aq)
(e) Hablur kuprum(II) nitrat berwarna biru terurai apabila dipanaskan untuk membentuk kuprum(II)
oksida yang hitam. Wasap coklat nitrogen dioksida dan gas oksigen juga terbebas.
Blue copper(II) nitrate crystals decomposes upon heating to form black copper(II) oxide. Brown fumes of
nitrogen dioxide and oxygen gas are also given off.
2Cu(NO3)2(p/s) → 2CuO(p/s) + 4NO2(g) + O2(g)
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Praktis Pentaksiran 3
Soalan Aneka Pilihan
Arahan: Pilih jawapan terbaik daripada pilihan A, B, C dan D.
Instructions: Choose the best answer from options A, B, C and D.
[Jisim atom relatif/Relative atomic mass: H = 1,
C = 12, O = 16, Cl = 35.5]
A 2.7 g O2
C 8.8 g CO2
D 71.0 g Cl2
B 6.2 g H2
1
Berapakah jisim karbon dioksida yang
mengandungi bilangan molekul yang sama
dengan 28 g nitrogen?
What is the mass of carbon dioxide that has the
same number of molecules as in 28 g nitrogen?
TP 1
4
Apakah data eksperimen yang diperlukan oleh
seorang pelajar untuk menentukan formula
empirik magnesium oksida?
What experimental data is required for a student
to determine the empirical formula of magnesium
oxide?
I Jisim magnesium dan oksigen yang
bergabung
Mass of magnesium and oxygen that
combined
II Isi padu gas oksigen yang digunakan
Volume of oxygen gas used
III Jisim atom relatif magnesium dan oksigen
Relative atomic mass of magnesium and
oxygen
IV Kedudukan logam magnesium dalam siri
kereaktifan
Position of magnesium in the reactivity series
C I, II dan III sahaja
A I dan III sahaja
I and III only I, II and III only
B II dan IV sahaja
D I, II, III, IV
II and IV only
[Jisim atom relatif/Relative atomic mass: C = 12,
N = 14; O = 16]
A 22 g
C 44 g
B 28 g
D 56 g
Mengaplikasi
TP 5
2
Berapakah bilangan molekul oksigen yang
mempunyai jisim yang sama dengan 64 g
kuprum(II) oksida?
What is the number of molecules of oxygen that
has the mass as 64 g copper(II) oxide?
TP 2
[Jisim atom relatif/Relative atomic mass: O
= 16, Cu = 64; Pemalar Avogadro/Avogadro
Constant = 6.0 x 1023 mol–1]
C 6.0 x 1023
A 4.8 x 1023
23
B 5.5 x 10
D 7.2 x 1024
3
Manakah antara bahan berikut mengandungi
bilangan atom paling banyak?
Which of the following substances has the most
number of atoms?
TP 3
Soalan Struktur
Arahan: Jawab semua soalan dalam bahagian ini.
Instructions: Answer all questions in this section.
1
Apabila dipanaskan dengan kuat, batu kapur terurai lalu membebaskan gas karbon dioksida. Dalam
suatu eksperimen, seorang pelajar memanaskan 5 g batu kapur dengan kuat sehingga jisimnya tidak
berubah lagi.
When heated strongly, limestone decomposes to produce carbon dioxide gas. In an experiment, a student
heated 5 g of limestone strongly until its mass does not change anymore. TP 4, 5
[Jisim atom relatif/Relative atomic mass: C = 12, O = 16, Ca = 40; Pemalar Avogadro/Avogadro constant =
6.0 x 1023 mol–1]
(a) Tuliskan persamaan bagi pemanasan batu kapur.
Write an equation for the heating of limestone.
CaCO3(p/s) → CaO(p/s) + CO2(g) [1 markah/mark]
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(b) (i) Apakah baki daripada pemanasan batu kapur?
What is the residue from the heating of limestone?
Kalsium oksida/Calcium oxide
[1 markah/mark]
(ii) Hitungkan jisim baki ini.
Calculate the mass of this residue.
Bilangan mol CaO/Number of moles of CaO = bilangan mol CaCO3/number of mols of CaCO3
= 5 ÷ [40 + 12 + 3(16)] = 0.05
Jisim CaO/Mass of CaO = 0.05[40 + 16] = 2.8 g
[1 markah/mark]
(c) Baki yang terhasil dalam tindak balas di atas larut sebahagian dalam air untuk membentuk air kapur.
Tulis persamaan kimia bagi tindak balas ini.
The above residue dissolves partially in water to form limewater. Write a chemical equation for the
reaction.
CaO(p/s) + H2O(ce/l) → Ca(OH)2(ak/aq) + CO2(g)
[1 markah/mark]
(d) Air kapur menjadi keruh apabila gas karbon dioksida dilalukan melaluinya.
Menilai
Limewater turns cloudy when carbon dioxide gas is passed through it.
(i) Terangkan pemerhatian di atas.
Explain the above observation.
Gas karbon dioksida bertindak balas dengan air kapur lalu membentukkan pepejal putih kalsium karbonat yang tidak larut.
Carbon dioxide gas reacts with limewater to form an insoluble white solid calcium carbonate.
[1 markah/mark]
(ii) Tuliskan persamaan kimia bagi tindak balas ini.
Write a chemical equation for this reaction.
CO2(g) + Ca(OH)2(ak/aq) → CaCO3(p/s) + H2O(ce/l) [1 markah/mark]
Soalan Esei
Arahan: Jawab semua soalan dalam bahagian ini.
Instructions: Answer all questions in this section.
1
(a) Terangkan maksud ‘formula kimia’. Nyatakan dua jenis maklumat yang boleh diperoleh daripada
formula kimia suatu sebatian.
Explain the meaning of ‘chemical formula’. State two types of information that can be obtained from the
chemical formula of a compound. TP 1, 3
[3 markah/marks]
(b) Komposisi mengikut peratusan jisim bagi asid suksinik ialah 40.7% karbon, 5.1% hidrogen dan
54.2% oksigen. Jika 0.59 g asid suksinik mengandungi 3.01 x 1021 molekul asid suksinik, tentukan
The percentage composition by mass of succinic acid is 40.7% carbon, 5.1% hydrogen and 54.2% oxygen.
If 0.59 g succinic acid contains 3.01 x 1021 succinic acid molecules, determine
(i) formula empirik asid suksinik
the empirical formula of succinic acid
[7 markah/marks]
(ii) jisim 1 mol asid suksinik
the mass of 1 mol of succinic acid
[2 markah/marks]
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(iii) formula molekul asid suksinik
the molecular formula of succinic acid
[Jisim atom relatif/Relative atomic mass: H = 1, C = 12, O = 16; Pemalar Avogadro/Avogadro
Constant = 6.0 x 1023 mol–1]
[2 markah/marks]
Genius
Menganalisis
SP 3.4.3
1
Hematit adalah bijih ferum(III) oksida. Logam ferum boleh diekstrak dengan memanaskan hematit
dengan karbon. Persamaan berikut menunjukkan tindak balas yang berlaku.
Haematite is an ore of iron(III) oxide. Iron can be extracted by heating haematite with carbon. The following
equation shows the reaction that happens. TP 4
Fe2O3(p/s) + C(p/s) → Fe(p/s) + CO2(g)
(a) Imbangkan persamaan di atas.
Balance the above equation.
2Fe2O3(p/s) + 3C(p/s) → 4Fe(p/s) + 3CO2(g)
(b) Jika 20 kg ferum(III) oksida dipanaskan dengan 20 kg carbon, hitungkan
If 20 kg iron(III) oxide is heated with 20 kg carbon, calculate
[Jisim atom relatif/Relative atomic mass: O = 16; Fe = 56; Isi padu molar gas/Molar volume of gas
Menganalisis
= 24 dm3]
(i) Jisim logam ferum yang didapati.
The mass of iron metal obtained.
Bilangan mol Fe2O3/Number of moles of Fe2O3 = = 125
Bilangan mol C/Number of moles of C = = 1666.7
[Karbon adalah berlebihan dan diabaikan semasa penghitungan/Carbon is in excess and is
ignored during the calculations]
Bilangan mol Fe = 2 x bilangan mol Fe2O3 = 2 x 125 = 250
Number of moles of Fe = 2 x number of moles of Fe2O3 = 2 x 125 = 250
Jisim/Mass of Fe = 250 x 56 = 14 000 g (14 kg)
(ii) Isipadu gas karbon dioksida yang terbebas pada keadaan bilik.
The volume of carbon dioxide gas produced at room conditions.
Mol CO2 = 1.5 x 125 = 187.5
Isi padu/Volume = 187.5 x 24 = 4500 dm3
Praktis Tambahan
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