Question 1:
a)
(1+1+2+3+3 = 10 marks)
Write Zr, V, and Mn in order of increasing atomic radius.
Answer: Mn < V < Zr
b)
(1 mark)
Write O, N, P in order of increasing ionization energy.
Answer: P < N < O
c)
(1 mark)
An atom has half as many protons as an atom of 28Si and also has six fewer neutrons than
an atom of 28Si. Give the symbol, including the mass number and the atomic
number, of this
atom.
Answer: 157N
e)
(2 mark)
Write (using a noble gas core) the electron configuration of Technetium (Tc) and Tc+2?
Which one is larger in size, Tc or Tc+2?
Answer:
Tc = [Kr] 5s24d5
(1 mark)
Tc is larger in size.
f)
Tc+2 = [Kr] 4d5
(1 mark)
(1 mark)
How many orbitals are there in the n=3 shell? Also write what are those orbitals.
Answer: Total 9 orbitals
(1 mark) and they are: one 3s orbital, three 3p orbitals, and five 3d
orbitals. (2 mark)
Question 2:
a)
(3+3+4 = 10 marks)
Calculate the pH of each of the following solutions and classify the solution as acidic
or basic:
i)
Urine: [H+] = 1.2 x 10-6 M
ii)
Blood: [H+] = 3.9 x 10-8 M
Answer:
Urine: [H+] = 1.2 x 10-6 M --- pH = -log10 [H+] = 5.92 acidic
(1.5 mark)
Blood: [H+] =3.9 x 10-8 M --- pH = -log10 [H+] = 7.41 basic
(1.5 mark)
b)
Write equations which represent the dissociation of H3AsO4 (a polyprotic acid) in
aqueous solution. Show each step of dissociation.
Answer:
Dissociation Step 1: H3AsO4 (aq) + H2O(l)⇌ H2AsO4- (aq) + H3O+(aq) (1 mark)
Dissociation Step 2: H2AsO4- (aq) + H2O(l)⇌ HAsO42- (aq) + H3O+(aq) (1 mark)
Dissociation Step 3: HAsO42- (aq) + H2O(l)⇌ AsO43- (aq) + H3O+(aq)
c)
(1 mark)
Write the acid-base reaction between CH3-CH2-COOH (acts as an acid) and
CH3MgBr. According to Brønsted–Lowry theory, identify each chemical as either an
‘acid’ or a ‘base’ and identify “conjugate” relationships.
Answer:
CH3-CH2-COOH + CH3MgBr  CH3-CH2-COOMgBr + CH4
(Acid)
(Base)
(Base)
(Acid)
(2 mark)
1ST Conjugate acid-base pair: CH3-CH2-COOH and CH3-CH2-COOMgBr (1 mark)
2ND Conjugate acid-base pair: CH4 and CH3MgBr
(1 mark)
Question 1:
(10 Marks)
(a) Between the elements, Nitrogen, Boron, Helium and Chlorine, which one is the most
reactive? Write the reason for your answer.
2 Marks
Ans: Chlorine. Reason: Chlorine only needs one more electron to complete its valence level,
making it extremely reactive and unstable. Nitrogen is reactive, but not too much (to gain 3
electrons). Boron is about as reactive as nitrogen (to lose 3 electrons), and helium is not
reactive at all, since it has a full two electrons in its valence level.
(b) Would you expect strontium to be, chemically, more similar to calcium or rubidium and
WHY?
Ans: Calcium. Reason: Both Ca and Sr have two valance electrons.
2 Marks
(c) What is the heaviest of the naturally occurring Noble gases? Write the name and the symbol.
1 Marks
Ans: Radon (Rn)
(d) What is the heaviest element with a one letter atomic symbol? Write the name and the
symbol.
1 Marks
Ans: Uranium (U)
(e) Among these ions of Nitrogen, which ion is largest in size? Provide a proper reason for your
answer.
2 Marks
N3-, N3+, N2- and N5+
Ans: N3- . Reason: Ions of atoms vary in their size. When an atom gains electrons the atomic
size increases, and when an atom loses electrons the atomic size decreases. Nitrogen with a
formal charge of negative three will have the largest size because it has the most electrons.
Remember that adding electrons will cause a negative ion, while removing electrons will
cause a positive ion.
(f) Among these groups in the periodic table, which group will have the lowest first ionization
energy? Provide a proper reason for your answer. 2 Marks
Group III, Halogens, Noble gases, Alkaline earth metals and Alkali metals.
Ans: Alkali Metals. Reason: Alkali metals only need to lose one electron in order to achieve
the stable noble gas octet. As a result, the loss of this electron requires very little energy
since the resulting ion is favorably stable. This property causes alkali metals (found in
group I) to have the lowest first ionization energies.
Question 2:
(10 Marks)
(a) In each case of the two reactions shown below, one of the reactants acts as an acid.
Identify it:
i) CH3COOH(aq) + HClO4(aq) == CH3COOH2+(aq) + ClO4-(aq)
ii) HCO3-(aq) + HSO4-(aq) == H2O(l) + CO2(g) + SO42-(aq)
Answer:
i) HClO4
ii) HSO4-
(1 mark)
(1 mark)
(b) Identify the acid-base conjugate pairs in the following reactions:
i) HCO3-(aq) + H3O+(aq) == CO2(g) + H2O(l) + H2O(l)
ii) H2SO4(aq) + HNO3(aq) == HSO4-(aq) + NO2+(aq) + H2O(l)
Answer:
i)
HCO3- (base) and CO2 + H2O (acid),
H3O+ (acid) and H2O (base)
(2marks)
ii)
H2SO4 (acid) and HSO4 (base),
HNO3 (base) and NO2+ + H2O (acid)
(2marks)
(c) Explain why in this process of electrolysis of water, the volume of gas collected over one
electrode double that of gas collected over the other electrode? What are these two gases?
(2marks)
Answer:
In water (H2O), hydrogen and oxygen are present in the ratio of 2:1 by volume. – 1 mark
(Also it can be checked with the balanced chemical equation of the electrolysis of water)
The two gases are Hydrogen (H2) and Oxygen (O2). – 1 mark
(d) Glacial acetic acid, pure HC2H3O2 (FW = 60.0), has a concentration of 17.54 M. If
85.5 mL of glacial acetic acid are diluted to 250.0 mL, what is the concentration of
the acetic acid now?
Answer: This is a dilution problem. Use M1V1 = M2V2
(2marks)
(17.54 M) (85.8 mL) = M2 (250.0 mL)
Thus, M2 = 5.998 M
Question 1:
a)
(3+2+2+1+1+1 = 10 marks)
Write an abbreviated (noble gas core) electron configuration for the following:
i)
Zn
ii)
O-2
iii)
Answer:
i)
Zn = [Ar] 4s2 3d10
(1 mark)
ii) O2- = [Ne]
(1 mark)
iii) Ni2+ = [Ar] 3d8
(1 mark)
Ni+2
b)
Which is the larger atom in each pair:
i)
Na or Si
ii)
P or Sb
Answer:
i) Na
c)
(1 mark)
ii) Sb
(1 mark)
Which atom has the larger ionization energy in each pair:
i)
B or O
ii)
Cl or Ar
Answer:
i) O
d)
(1 mark)
ii) Cl
Write Sc, Sr, and Sc+3 in order of increasing atomic radius.
Answer: Sc3+ < Sc < Sr
e)
(1 mark)
Write Al, In, and P in order of increasing ionization energy.
Answer: In < Al < P
f)
(1 mark)
(1 mark)
Which noble gas that does not have the ns2np6 type of valance shell electron
configuration?
Answer: He (1s2) (1 mark)
Question 2:
a)
(4+2+4 = 10 marks)
For a 7.4 x 10-3 M HCl solution, determine the following:
i)
[H+]
ii)
[OH-]
iii)
the pH
Answer:
i) Because HCl is a strong acid, it completely dissociates:
HCl  H+ + ClSo the concentration of hydrogen ion of a 7.4 x 10-3 M HCl solution is 7.4 x 10-3 M.
[H+] = 7.4 x 10-3
(1 mark)
ii) Substituting the [H+] value in Kw = [H+] x [OH-] = 1.0 x10-14
(1/2 mark)
------ [OH-] = 7.4 x 10-3 (1 mark)
iii) pH = -log10 [H+]
b)
(1/2 mark) --- pH = -log(7.4 x 10-3) ---- pH = 2.13 (1 mark)
Write equations which represent the dissociation of C6H5NH2 (a weak base) in
aqueous solution.
Answer:
C6H5NH2 (aq) + H2O (l) ⇌ C6H5NH3+ (aq) + OH- (aq)
(2 mark)
Write the acid-base reaction between CH3-CH2-NH3+ (acts as an acid) and CH3OH.
c)
According to Brønsted–Lowry theory, identify each chemical as either an ‘acid’ or a ‘base’ and
identify “conjugate” relationships.
Answer:
CH3-CH2-NH3+ + CH3OH ⇌ CH3-CH2-NH2 + CH3OH2+
(Acid)
(Base)
(Base)
(2 mark)
(Acid)
1ST Conjugate acid-base pair: CH3-CH2-NH3+ and CH3-CH2-NH2
2ND Conjugate acid-base pair: CH3OH and CH3OH2+
(1 mark)
(1 mark)
QUESTION 1
1.1
[16]
Perform the following calculations and give the answer to the correct number of significant figures.
a.
(0.871 x 0.57) ÷ 5.971
(2)
= 0.49647 + 5.971 
= 6.467
= 6.47 
b.
0.00015 x 54.6 + 1.032
(2)
= 0.00819 + 1.032 
= 1.04
= 1.0 
1.2
A submicroscopic particle suspended in a solution has a volume of 1.3 µm3. Convert this volume to
liters.
(3)
= 1.3 x (1m/106)3 x (10-1/1m)3 
= 1.3 x 10-21 dm3 
1 dm3 = 1 L 
Therefore = 1.3 x 10-21L 
1.3
Iodine is a bluish-black solid. It forms a violet-colored vapor when heated. The solid melts at 235°F.
What is this temperature in degrees Celsius and in Kelvin?
°C = (°F x 1.8) – 32 
(4)
= 235 *1.8 – 32
= 113 C ° 
K = °C + 273.15 
K = 386.15 
1.4
The oxygen molecule consists of two oxygen atoms a distance of 121 pm apart. How many
millimeters is this distance?
(2)
= 121 pm x (1m/1012pm)  x (103mm/1mm) 
= 1.21 x 10-7 mm 
1.5
Iron wire weighing 5.6 g is placed in a beaker and covered with 15.0 g of dilute hydrochloric acid.
The acid reacts with the metal and gives off hydrogen gas, which escapes into the surrounding air.
After reaction, the contents of the beaker weigh 20.4 g. What is the mass of the hydrogen gas
produced by the reaction?
(3)
By Law of Conservation of Mass 
Fe
+
5.6 g
2HCl

H2(g)
15 g
+
FeCl2(s)
x
20.4 g
Mass of Hydrogen = 20.4 – (5.6 + 15)  = 0.2 g 
QUESTION 2
2.1
2.2
[4]
Indicate whether each of the following is a physical or a chemical property of the substance:
a.
Sodium chloride dissolves in water. Physical
(1)
b.
Calcium carbonate fizzes when mixed with vinegar. Chemical
(1)
Which of the following are pure substances and which are mixtures? For each, list all the
different phases present:
a.
Mercury liquid and its vapor. Pure substance
(1)
b.
Paint, containing a liquid solution and a dispersed solid pigment.
(1)
Heterogeneous mixture
QUESTION 3
3.1
3.2
[6]
Name the following compounds:
(a)
Mg(HCO3)2
Magnesium hydrogen carbonate or Magnesium bicarbonate (1)
(b)
Hg2(CH3COO-)2 Mercury (I) acetate
Give the molecular formula of each of the following compounds:
(1)
3.3
(a)
Praseodymium(III) chloride heptahydrate
(b)
Antimony(V) sulfide
PrCl3•7H2O
(1)
Sb2S5
(1)
How many rhodium cations are present in a 10.0 g sample of rhodium(III) sulfate? Show all your
calculations.
(2)
Rhodium(III) sulfate = Rh2(SO4)3 = 494.01 g/mol
10.0 g/494.10 g/mol = 0.020242505 moles
1 mol of Rh2(SO4)3 contains 2×6.022×1023 Rh3+ cations
Therefore, 0.020242505 moles contain: 0.020242505×2×6.022×1023
= 2.44×1022 Rh3+ cations
QUESTION 4
[6]
Epsom salts is a hydrated ionic compound with the following formula: MgSO4•xH2O. It takes its name from
a bitter saline spring in Epsom in Surrey, England, where the salt was produced from the springs. One of
its common uses is as an oral laxative. A 4.93 g sample of Epsom salts was heated to drive off the water
of hydration. The mass of the sample after complete dehydration was 2.41 g.
Find the number of waters of hydration (the value of x) in Epsom salts.
(6)
4.93 g of MgSO4•xH2O gave 2.41 g of MgSO4 and (4.93 – 2.41 =) 2.52 g H2O
Molar mass of MgSO4 = 120.371 g/mol
2.41 g MgSO4 = 0.02002 moles
Molar mass of H2O = 18.016 g/mol
2.52 g H2O = 0.139876 mole
This means 0.02002 moles MgSO4 carries 0.139876 moles of water of crystallization
1.0 mole MgSO4 carries x moles of water of crystallization
x = 0.139876 ×1.0/0.02002 = 6.9867 ≈ 7
Therefore the formula of the hydrate is MgSO4•7H2O
QUESTION 5
[8]
Lead is found in Earth’s crust as several different lead ores mainly Galena (PbS), Cerussite (PbCO 3) and
Anglesite (PbSO4).
5.1
Which of the three ores above contains more lead by mass? Please show all your calculations.
(3)
%Mass of Pb in Galena (PbS): 207.2/(207.2 + 32.07) × 100% = 86.6%
%Mass of Pb in Cerussite (PbCO3): 207.2/[207.2 + 12.01 + (16.00×3)] × 100% = 77.5%
%Mass of Pb in Anglesite (PbSO4): 207.2/[207.2 + 32.07 + (16.00×4)] × 100% = 68.3%
5.2
Suppose a certain rock contains 38.0% PbS, 25.0% PbCO3 and 17.4% PbSO4, with the remainder of
the rock being substances that do not contain lead. How much of this rock, in kilograms, must be
processed to obtain 5.0 tons of lead?
(5)
Total Percentage by Mass of Lead in the rock from all three minerals:
PbS: (38%×86.6%) + PbCO3: 25%×77.5% + 17.4%×68.3% = 64.2%
This means 1000 g of this rock contains 642 g Pb or 1.0 kg contains 0.642 kg Pb
Thus, 5 tons = 5000 kg will be contained in:
5000/0.642 = 7788.2 kg = 7.79 tons of the rock.
QUESTION 6
6.1
[10]
Write Zr, V, and Mn in order of increasing atomic radius.
(1)
Mn < V < Zr
6.2
Write O, N, P in order of increasing ionization energy.
(1)
P<N<O
6.3
An atom has half as many protons as an atom of 28Si and also has six fewer neutrons than an atom
of 28Si. Give the symbol, including the mass number and the atomic number, of this atom.
(2)
15
6.4
7N
Write (using a noble gas core) the electron configuration of Technetium (Tc) and Tc2+? Which one
is larger in size, Tc or Tc2+?
Tc = [Kr] 5s24d5 (1 mark)
(3)
Tc+2 = [Kr] 4d5
(1 mark)
Tc is larger in size. (1 mark)
6.5
How many orbitals are there in the n=3 shell? Also write what are those orbitals.
(3)
9 orbitals (1 mark) and they are: one 3s orbital, three 3p orbitals, and five 3d orbitals. (2 marks)
QUESTION 7
7.1
[10]
Calculate the pH of each of the following solutions and classify the solution as acidic
or
basic:
i)
Urine: [H+] = 1.2 x 10-6 M
(1.5)
Urine: [H+] = 1.2 x 10-6 M --- pH = -log10 [H+] = 5.92 (acidic)
ii)
Blood: [H+] = 3.9 x 10-8 M
Blood: [H+] =3.9 x 10-8 M --- pH = -log10 [H+] = 7.41 (basic)
(1.5)
7.2
Write equations which represent the dissociation of H3AsO4 (a polyprotic acid) in
aqueous solution. Show each step of dissociation.
(3)
Dissociation Step 1: H3AsO4(aq) + H2O(l) ⇌ H2AsO4-(aq) + H3O+(aq) (1 mark)
Dissociation Step 2: H2AsO4-(aq) + H2O(l) ⇌ HAsO42-(aq) + H3O+(aq) (1 mark)
Dissociation Step 3: HAsO42-(aq) + H2O(l) ⇌ AsO43-(aq) + H3O+(aq) (1 mark)
7.3
Write the acid-base reaction between CH3-CH2-COOH (acts as an acid) and
According to Brønsted–Lowry theory, identify each chemical as either an
identify “conjugate” relationships.
CH3MgBr.
‘acid’ or a ‘base’ and
(4)
CH3-CH2-COOH + CH3MgBr  CH3-CH2-COOMgBr + CH4
(Acid)
(Base)
(Base)
(Acid)
(2 marks)
1ST Conjugate acid-base pair: CH3-CH2-COOH and CH3-CH2-COOMgBr (1 mark)
2ND Conjugate acid-base pair: CH4 and CH3MgBr .(1 mark)
QUESTION 8
8.1
[12]
Write a balanced equation for the reaction of solid aluminum with aqueous sulfuric acid to form
aqueous aluminum sulfate and hydrogen gas.
(5)
2Al(s) + 3H2SO4(aq)  Al2(SO4)3(aq) + 3H2(g)
8.2
Consider the following reaction occurring in aqueous solution.
2 HBr(aq) + Ca(OH)2(aq)  2 H2O(l) + CaBr2(aq)
Write a complete ionic equation and net ionic equation for the reaction.
(2)
2H+ (aq) + 2Br- (aq) + Ca2+ (aq) + 2OH- (aq)  2H2O (l) + Ca2+ (aq) + 2Br- (aq)
H+ (aq) + OH- (aq)  H2O (l)
8.3
Classify each reaction as a synthesis, decomposition, single-displacement, or doubledisplacement.
i. 2Al(s) + 2H3PO4(aq)  2AlPO4(aq) + 3 H2(g) Single-displacement
(1)
ii. CuSO4(aq) + 2 KOH(aq)  Cu(OH)2(s) K2SO4(aq) Double displacement
(1)
iii. CuCl6(aq)
8.4
Electric current
Cu(s) + Cl2(g) Decomposition
Which of the reactions shown in 8.3 above are redox reactions? i. and iii.
QUESTION 9
(1)
(2)
[9]
9.1
The following reaction is used to obtain iron from iron ore:
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
The reaction of 185 g of Fe2O3 with 95.3 g of CO produces 87.4 g of Fe. Find
i. the limiting reactant, CO
(4)
ii. theoretical yield, and 126.7 g Fe
(1)
iii. percent yield.
(1)
68.98% ~ 69% yield
185 𝑔 𝐹𝑒2𝑂3 𝑥
1 𝑚𝑜𝑙 𝐹𝑒2𝑂3
2 𝑚𝑜𝑙 𝐹𝑒
55.85 𝑔 𝐹𝑒
𝑥
𝑥
= 129.4 𝑔 𝐹𝑒
159.7 𝑔 𝐹𝑒2𝑂3 1 𝑚𝑜𝑙 𝐹𝑒2𝑂3
1 𝑚𝑜𝑙 𝐹𝑒
95.3 𝑔 𝐶𝑂 𝑥
1 𝑚𝑜𝑙 𝐶𝑂
2 𝑚𝑜𝑙 𝐹𝑒
55.85 𝑔 𝐹𝑒
𝑥
𝑥
= 126.7 𝑔 𝐹𝑒
28.01 𝑔 𝐶𝑂 3 𝑚𝑜𝑙 𝐶𝑂
1 𝑚𝑜𝑙 𝐹𝑒
87.4 𝑔 𝐹𝑒
𝑥 100% = 69%
126.7 𝑔 𝐹𝑒
9.2
Describe how you would make 500.0 mL of a 0.200 M NaOH solution from a 15.0 M stock NaOH
solution.
(3)
Given M1 = 15.0 M
Find V1 and determine how much distilled water need to be added.
M2 = 0.200 M
V2 = 500.0 mL
From M1V1=M2V2
𝑉1 =
𝑀2𝑉2
(0.200 𝑀)(500.0 𝑚𝐿)
=
= 6.67 𝑚𝐿
𝑀1
15.0 𝑀
Therefore take 6.67 mL of the stock solution in a 500.0 mL volumetric flask and add 493.33 mL of
deionized water.
QUESTION 1
1.1
[12]
Perform the following calculations and give the answer to the correct number of significant figures.
a.
(0.311 x 0.57) ÷ 5.93
(2)
= 0.17727  ÷ 5.93
= 0.02989
= 0.030
b.
0.0015 x 554.6 + 1.032
= (0.8319 x 554.6) + 1.032
= 1.863
= 1.86
(2)
1.2
A submicroscopic particle suspended in a solution has a volume of 2.8 nm3. Convert this volume to
liters.
(3)
2.8 x (1m/109)3 x (10-1/1m)3 
= 2.8 x 10-24 dm3 
1 dm3 = 1 L 
Therefore = 2.8 x 10-24L 
1.3
One litre of gasoline in an automobile’s engine produces on the average 9.5 kg of carbon dioxide,
which is a greenhouse gas, that is, it promotes the warming of Earth’s atmosphere. Calculate the
annual production of carbon dioxide in kilograms and µg if there are 40 million cars in the South
Africa and each car covers a distance of 5000 km at a consumption rate of 20 km per litre.
(5)
40 x 106 x (5000 km/car)  x (1 litre/20 km)  x (9.5 kg/1L) 
= 9.5 x 1010 kg CO2
In μg
= 9.5 x 1010 kg x (103 g/ 1kg)  x (106 ug/1 g)  = 9.5 x 1019 μg 
QUESTION 2
[8]
When 20.0 g of A was heated, 8.8 g of B was given off, leaving 10.2 g of E. The same quantity of B can
also be prepared by a combination of 2.4 g of C and 3.2 g of D. The E can be electrolysed, after melting,
to yield 8.0 g of F and 1.6 g of G, neither of which can be further decomposed by ordinary chemical means.
Identify the lettered materials A, B, E, F and G as either an element or a compound and explain your
decision.
A + E  can be decomposed   compound 
B is a combination of C and D  compound 
F + G  can be broken down   elements 
QUESTION 3
3.1
[4]
Name the following compounds:
(a)
VSO4•7H2O
(b)
SF6
Vanadium(II) sulfate heptahydrate
Sulfur hexafluoride
(1)
(1)
3.2
Give the molecular formula of each of the following compounds:
(a)
Tin(IV) chromate
Sn(CrO4)2
(1)
(b)
Thorium dicarbide
ThC2
(1)
QUESTION 4
[6]
A white powder suspected to be the illegal narcotic cocaine, is subjected to combustion analysis at the
SAPS Forensic Science Laboratories and gave the following results:
C, 67.31%; H, 6.98%; O, 21.10% and the rest is Nitrogen.
Further analysis using a variety of scientific instruments confirmed the powder to be pure cocaine, with a
molecular mass of 303.35294. Please work out the molecular formula of cocaine.
100 g of the powder contains 67.3 g C; 6.98 g H; 21.10 g O and 4.62 g N
Number of moles:
Mole ratios
C = 67.31/12.01 = 5.604496253
5.604496253/0.329764454 = 16.995 = 17
H = 6.98/1.008 = 6.924603175
6.924603175/0.329764454 = 20.999 = 21
O = 21.10/16.00 = 1.31875
1.31875/0.329764454 = 3.999 = 4
N = 4.62/14.01 = 0.329764454
0.329764454/0.329764454 = 1
Molecular formula: C17H21NO4 which is consistent with the molecular mass 303.35294
QUESTION 5
[10]
Rare-earth metals (REM), as defined by IUPAC, are the fifteen lanthanides in the Periodic Table, as well as
scandium and yttrium. The rare-earth elements are often found together in several minerals. Examples of
such minerals are Gadolinite with the composition (Ce,La,Nd,Y)2FeBe2Si2O10, Euxenite with the formula
(Y,Ca,Ce,U,Th)(Ti,Nb,Ta)2O6, and Blomstrandine with the formula (Y,Ca,Fe,Th)(Ti,Nb)2(O,OH)6.
5.1
Which of the three minerals above contains more ytterbium by mass? Please show all your
calculations.
(5)
Percentage mass of Y in Gadolinite (Ce,La,Nd,Y)2FeBe2Si2O10:
(2×88.91)/{[(140.1+138.9+144.2+88.91)×2]+55.85+(9.012×2)+(28.09×2)+(16.00×10)]} × 100%
= 13.53%
Percentage mass of Y in Euxenite (Y,Ca,Ce,U,Th)(Ti,Nb,Ta)2O6:
88.91/{88.91+40.08+140.1+238.0+232.0+[(47.88+92.91+180.9)×2]+(16.00×6)} × 100%
= 6.01%
Percentage mass of Y in Blomstrandine (Y,Ca,Fe,Th)(Ti,Nb)2(O,OH)6:
88.91/{88.91+40.08+55.85+232+(47.88+92.91)×2 +[(16.00+(16.00+1.008)]×6} × 100%
= 9.92%
5.2
What mass of cerium, in kilograms, can be obtained from 1 ton of gadolinite?
(5)
Percentage mass of Ce in Gadolinite (Ce,La,Nd,Y)2FeBe2Si2O10:
140.1/{[140.1+138.9+144.2+88.91)×2]+55.85+(9.012×2)+(28.09×2)+(16.00×10)]} × 100%
= 10.66%
In 1000 kg of gadolinite there will be 1000 kg × 0.1066 = 106.6 kg
QUESTION 6
6.1
[10]
Between the elements, Nitrogen, Boron, Helium and Chlorine, which one is the most reactive?
Write the reason for your answer.
(2)
Chlorine. Chlorine only needs one more electron to complete its valence level, making it extremely
reactive and unstable. Nitrogen is reactive, but not too much (to gain 3 electrons). Boron is about
as reactive as nitrogen (to lose 3 electrons), and helium is not reactive at all, since it has a full two
electrons in its valence level.
6.2
Would you expect strontium to be, chemically, more similar to calcium or rubidium and WHY?
Calcium. Both Ca and Sr have two valance electrons.
6.3
What is the heaviest of the naturally occurring Noble gases? Write the name and the symbol.
(1)
Radon (Rn)
6.4
What is the heaviest element with a one letter atomic symbol? Write the name and the symbol.
(1)
Uranium (U)
6.5
Among these ions of Nitrogen, which ion is largest in size? Provide a proper reason for your
answer.
(2)
N3-, N3+, N2- and N5+
N3- . Reason: Ions of atoms vary in their size. When an atom gains electrons the atomic size
increases, and when an atom loses electrons the atomic size decreases. Nitrogen with a formal
charge of negative three will have the largest size because it has the most electrons. Remember
that adding electrons will cause a negative ion, while removing electrons will cause a positive ion.
6.6
Among these groups in the Periodic Table, which group will have the lowest first ionization
energy? Provide a proper reason for your answer.
(2)
Group III, Halogens, Noble gases, Alkaline earth metals and Alkali metals.
Alkali Metals. Alkali metals only need to lose one electron in order to achieve the stable noble gas
octet. As a result, the loss of this electron requires very little energy since the resulting ion is
favorably stable. This property causes alkali metals (found in group I) to have the lowest first
ionization energies.
QUESTION 7
[10]
7.1 In each case of the two reactions shown below, one of the reactants acts as an acid. Identify it:
CH3COOH2+(aq) + ClO4-(aq)
iii) CH3COOH(aq) + HClO4(aq)
iv) HCO3-(aq) + HSO4-(aq)
(1)
H2O(l) + CO2(g) + SO42-(aq)
(1)
7.2 Identify the acid-base conjugate pairs in the following reactions:
iii) HCO3-(aq) + H3O+(aq)
CO2(g) + H2O(l) + H2O(l)
(2)
HCO3- (base) and CO2 + H2O (acid),
H3O+ (acid) and H2O (base)
iv) H2SO4(aq) + HNO3(aq)
HSO4-(aq) + NO2+(aq) + H2O(l)
(2)
H2SO4 (acid) and HSO4- (base),
HNO3 (base) and NO2+ + H2O (acid)
7.3 Explain why in this process of electrolysis of water, the volume of gas collected over one electrode
double that of gas collected over the other electrode? What are these two gases?
(2)
In water (H2O), hydrogen and oxygen are present in the ratio of 2:1 by volume. – 1 mark
(Also it can be checked with the balanced chemical equation of the electrolysis of water)
The two gases are Hydrogen (H2) and Oxygen (O2). – 1 mark
7.4 Glacial acetic acid, pure CH3COOH (FW = 60.0), has a concentration of 17.54 M. If 85.5 mL of glacial
acetic acid are diluted to 250.0 mL, what is the concentration of the acetic acid now?
(2)
This is a dilution problem. Use M1V1 = M2V2 (2marks)
(17.54 M) (85.8 mL) = M2 (250.0 mL)
Thus, M2 = 5.998 M
QUESTION 8
8.1
[9]
Consider the following unbalanced reactions occurring in aqueous solutions and write a
complete balanced ionic equation and a net ionic equation for each reaction:
i. Pb(NO3)2(aq) + LiCl(aq)
PbCl2(s) + LiNO3(aq)
(2)
Pb2+(aq) + 2NO3-(aq) + 2Li+(aq) + 2Cl-(aq)  PbCl2(s) + 2Li+(aq) + 2NO3-(aq)
Pb2+(aq) + 2Cl- (aq)  PbCl2 (s)
ii. HBr(aq) + Ca(OH)2(aq)
H2O(l) + CaBr2(aq)
(2)
2H+(aq) + 2Br-(aq) + Ca2+(aq) + 2OH-(aq)  2H2O(l) + Ca2+(aq) + 2Br-(aq)
H+(aq) + OH-(aq)  H2O(l)
8.2
Classify each reaction as a synthesis, decomposition, single-displacement, or doubledisplacement. 1 mark each
i. Na2O(s) + H2O(l)
2 NaOH(aq) Synthesis
ii. Ba(NO3)2(aq) + K2SO4(aq)
BaSO4(s) + KNO3(aq) Double displacement
iii. 2Al(s) + Fe2O3(s)
8.3
Al2O3(s) + O2(g) Single-displacement
Which of the reactions shown in 8.2 above are redox reactions? iii.
QUESTION 9
9.1
(1)
(1)
(1)
(2)
[12]
Consider the reaction between calcium oxide and carbon dioxide
CaO(s) + CO2(g)
(6)
CaCO3(s)
Bongani allows 14.4 g of CaO and 13.8 g of CO2 to react. When the reaction is finished, he
collects 19.4 g of CaCO3. Find
i. the limiting reactant, CaO
ii. theoretical yield, 25.7 g CaCO3
iii. percent yield. 75.5%
14.4 𝑔 𝐶𝑎𝑂 𝑥
1 𝑚𝑜𝑙 𝐶𝑎𝑂
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3 100.09 𝑔 𝐶𝑎𝐶𝑂3
𝑥
𝑥
= 25.7 𝑔 𝐶𝑎𝐶𝑂3
56.08 𝑔 𝐶𝑎𝑂
1 𝑚𝑜𝑙 𝐶𝑎𝑂
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3
13.8 𝑔 𝐶𝑂2 𝑥
1 𝑚𝑜𝑙 𝐶𝑂2
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3 100.09 𝑔 𝐶𝑎𝐶𝑂3
𝑥
𝑥
= 31.4 𝑔 𝐶𝑎𝐶𝑂3
44.01 𝑔 𝐶𝑂2
1 𝑚𝑜𝑙 𝐶𝑂2
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3
𝑦𝑖𝑒𝑙𝑑% =
9.2
19.4 𝑔 𝐶𝑎𝐶𝑂3
𝑥 100% = 75.5%
25.7 𝑔 𝐶𝑎𝐶𝑂3
How much of a 1.25 M sodium chloride solution in milliliters is required to completely
precipitate all of the silver in 25.0 mL of a 0.45 M silver nitrate solution.
(3)
NaCl(aq) + AgNO3 (aq)  AgCl (s) + NaNO3
Given M1 = 1.25 M
Find V1
M2 = 0.45 M
V2 = 25.0 mL
M1V1=M2V2
𝑉1 =
9.3
𝑀2𝑉2
𝑀1
=
(0.45 𝑀)(25.0 𝑚𝐿)
1.25 𝑀
= 9.0 𝑚𝐿
A laboratory procedure calls for making 500.0 mL of a 1.4 M KNO3 solution. How much KNO3 in
grams is needed?
𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
(3)
𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠
𝑉𝑜𝑙.𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
𝑁𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑥 𝑉𝑜𝑙. 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
No. of moles of KNO3 = 1.4 mol/L x 500.0 mL = 700 mMol = 0.700 mol
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑚𝑎𝑠𝑠
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
Actual mass = No of Moles x Molar mass = 0.700 mol KNO3 x 101.11 g/mol = 70.78 g KNO3
QUESTION 1
1.1
1.2
[9]
How many significant figures are there in the following measurements?
(a) 0.770090 m 6 sig figs
(1)
(b) 4.1003 × 104 cm
(1)
5 sig figs
Perform the following operations and express the answers with the appropriate number of
significant figures.
(a) 320.5 – (6104.5/2.3)
(2)
= 320.5 – 2654.1304 
= - 2.3 x 103 
(b) [258.3 x 105) – (1.200 x 103)] x 2.8954
(2)
= 25828800 x 2.8954 
= 7.478 x 107 
1.3
Pure water freezes at 32 °F. If 10% by mass NaCl is added to the water, the freezing point drops
to about 20.3 °F. What is this difference expressed in °C
(3)
Difference = 32 – 20.3 = 11.3 °F 
°C = (°F x 1.8) – 32 
= 11.7 *1.8 – 32
= - 25.5 C ° 
QUESTION 2
2.1
[11]
A 33.0-g sample of an unknown liquid at 20.0 °C is heated to120 °C. During this heating, the density
of the liquid changes from 0.854 g/cm3 to 0.797 g/cm3. What volume would this sample occupy
at 120 °C?
(3)
V =
2.2
33.0 g
m
 =
= 41.405 = 41.4 cm3 
0.797 g/cm3
d
What is the volume in cubic millimeters of a 3.54 carat diamond, given that the density of diamond
is 3.51 g/mL? (1 carat = 200 mg). Assuming that the diamond is generally spherical (volume of a
sphere is V =
4 3
pr , what is the approximate diameter of the diamond in mm?
3
(8)
Vol of diamond = 3.45 carat (200/1carat)  (1g/1000mg)  (1cm3/3.51 g) 
= 0.197 cm3 
V = 4/3πr3 
0.197 = 4/3 πr3 
r = 0.361 cm 
2r = 0.722 mm x 10mm/1cm = 7.22 mm 
QUESTION 3
3.1
3.2
[6]
Name the following compounds:
(a)
(NH4)2CO3
(b)
NiB
(Ammonium carbonate)
(Nickel(III) boride)
Give the molecular formula of each of the following compounds:
(1)
(1)
3.3
(a)
Lutetium(III) sulfate octahydrate
(b)
Tungsten(II) carbide
(Lu2(SO4)3•8H2O)
(1)
(W2C)
(1)
How many thallium cations are present in a 10.0 g sample of thallium(III) nitrate? Show all your
calculations.
(2)
Tl(NO3)3 = 328.42 g/mol
10.0 g/328.42 g/mol = 0.030448815 moles
1 mole Tl(NO3)3 contains 6.022 × 1023 Tl3+ cations
Therefore, 0.030448815 moles contain 1.834 × 1023 Tl3+ cations.
QUESTION 4
[5]
Ascorbic acid (vitamin C) functions as a cofactor in many enzymatic reactions in animals (and humans)
that mediate a variety of essential biological functions, including wound healing and collagen synthesis. It
is composed of 40.92% by mass Carbon, 54.50% Oxygen and 4.58% Hydrogen. If the molecular weight of
vitamin C is 176.12 amu, determine the molecular formula of ascorbic acid.
(5)
100 grams of ascorbic acid contain: 40.92 g C; 54.50 g O and 4.58 g H
40.92 g/12.01 g/mol = 3.407160699 mol Carbon;
54.50 g/16.00 g/mol = 3.40625 mol Oxygen;
4.58 g/1.008 g/mol = 4.543650794 mol Hydrogen.
Dividing throughout by the smallest number of moles gives C1.000 H1.33915829 O1.000
Multiplying by the smallest whole number (3) to get whole number ratios gives the Empirical formula
C3H4O3
Molar mass of C3H4O3 = 88.062 = 176.124/2
The Molecular Formula of ascorbic acid is thus (C3H4O3)2 = C6H8O6
QUESTION 5
[9]
In nature, copper occurs in a variety of minerals as sulfides, sulfosalts, carbonates and oxides. Examples
of copper containing minerals are: chalcopyrite [CuFeS2], enargite [Cu3AsS4], and malachite
[Cu2CO3(OH)2].
5.1
Which of the three ores above contains more copper by mass? Please show all your calculations.
(4.5)
Mass% of Cu in chalcopyrite [CuFeS2]:
63.55/(63.55 + 55.85 + 32.07×2) × 100% = 34.62%
Mass% of Cu in enargite [Cu3AsS4]:
(3×63.55)/(3×63.55 + 74.92 + 32.07×4) × 100% = 48.41%
Mass% of Cu in malachite Cu2CO3(OH)2]:
(2×63.55)/(2×63.55 + 12.01 + 1.008×2 + 16.00×5) × 100% = 57.48%
5.2
Suppose a certain rock contains 0.75% CuO. What mass of this rock, in tons, must be processed in
order to obtain 750 kilograms of pure copper?
(4.5)
The rock contains 0.75% CuO.
The mass percent of copper in CuO = 63.55/(63.55 + 16.00) × 100% = 79.8868636%
The mass percent of copper in the rock is thus 0.75 × 0.798868636 = 0.599%
The mass of rock to be processed = 750 kg/0.599 = 12520.87 kg = 12.52 tons.
QUESTION 6
6.1
6.2
6.3
[10]
Write an abbreviated (noble gas core) electron configuration for the following:
(a)
Zn
[Ar] 4s2 3d10
(1)
(b)
O-2
[Ne]
(1)
(c)
Ni+2
[Ar] 3d8
(1)
Which is the larger atom in each pair:
(a)
Na or Si
Na
(1)
(b)
P or Sb
Sb
(1)
Which atom has the larger ionization energy in each pair?
(a)
B or O
O
(1)
(b)
Cl or Ar
Ar
(1)
6.4
Write Sc, Sr, and Sc+3 in order of increasing atomic radius. Sc3+ < Sc < Sr
(1)
6.5
Write Al, In, and P in order of increasing ionization energy. In < Al < P
(1)
6.6
Which noble gas that does not have the ns2np6 type of valance shell electron
He = (1s2)
QUESTION 7
configuration?
(1)
[10]
7.1
For a 7.4 x 10-3 M HCl solution, determine the following:
(a)
[H+]
(1 mark)
Because HCl is a strong acid, it completely dissociates: HCl  H+ + ClThe concentration of hydrogen ion [H+] of a 7.4 x 10-3 M HCl solution is thus 7.4 x 10-3 M.
(b)
[OH-]
(1.5 marks)
Substituting the [H+] value in Kw = [H+] x [OH-] = 1.0 x10-14
[OH-] = 7.4 x 10-3
(c)
pH
(1.5 marks)
pH = -log10 [H+] = -log(7.4 x 10-3) = 2.13
7.2
Write equations which represent the dissociation of C6H5NH2 (a weak base) in
solution.
aqueous
(2)
C6H5NH2 (aq) + H2O (l) ⇌ C6H5NH3+ (aq) + OH- (aq)
7.3
Write the acid-base reaction between CH3-CH2-NH3+ (acts as an acid) and CH3OH. According to
Brønsted–Lowry theory, identify each chemical as either an ‘acid’ or a ‘base’ and identify
“conjugate” relationships.
CH3-CH2-NH3+
(Acid)
+
(4)
⇌
CH3OH
CH3-CH2-NH2
(Base)
+
(Base)
CH3OH2+
(Acid)
(2 marks)
1ST Conjugate acid-base pair: CH3-CH2-NH3+ and CH3-CH2-NH2
(1 mark)
2ND Conjugate acid-base pair: CH3OH and CH3OH2+
(1 mark)
QUESTION 8
8.1
[22]
Write balanced complete ionic and net ionic equations for each of the following reactions:
(a) AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq)
(2)
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)  AgCl(s) + K+ (aq) + NO3- (aq)
Ag+ (aq) + Cl- (aq)  AgCl(s)
(b) NaOH(aq) + HNO3(aq)  H2O(l) + NaNO3(aq)
(2)
Na+ (aq) + OH- (aq) + H+ (aq) + NO3- (aq)  H2O (l) + Na+ (aq) + NO3- (aq)
OH- (aq) + H+ (aq)  H2O (l)
8.2
Balance each of the following chemical reactions and classify each as a synthesis, decomposition,
single-displacement, or double-displacement reaction:
(a) K2S(aq) + Co(NO3)2(aq)  KNO3(aq) + CoS(s)
(b) H2(g) + N2(g)  NH3(g)
Synthesis
Double-displacement
(1)
(1)
(c) Zn(s) + CoCl2(aq)  ZnCl2(aq) + Co(s)
Single-displacement
(1)
8.3
Which of the above reactions are redox reactions? (b) and (c)
(2)
8.4
Magnesium oxide can be produced by heating magnesium metal in the presence of oxygen. The
balanced equation for the reaction is:
2Mg(s) + O2(g)  2MgO(s)
When 10.1 g of Mg react with 10.5 g of O2, 11.9 g of MgO are collected. Determine the limiting
reactant, theoretical yield and percent yield for the reaction.
10.1 𝑔 𝑀𝑔 𝑥
10.5 𝑔 𝑂2 𝑥
(5)
1 𝑚𝑜𝑙 𝑀𝑔
2 𝑚𝑜𝑙 𝑀𝑔𝑂 40.31 𝑔 𝑀𝑔𝑂
𝑥
𝑥
= 16.8 𝑔 𝑀𝑔𝑂
24.31 𝑔 𝑀𝑔
2 𝑚𝑜𝑙 𝑀𝑔
1 𝑚𝑜𝑙 𝑀𝑔𝑂
1 𝑚𝑜𝑙 𝑂2 2 𝑚𝑜𝑙 𝑀𝑔𝑂 40.31 𝑔 𝑀𝑔𝑂
𝑥
𝑥
= 26.4 𝑔 𝑀𝑔𝑂
32 𝑔 𝑂2
1 𝑚𝑜𝑙 𝑂2
1 𝑚𝑜𝑙 𝑀𝑔𝑂
𝑦𝑖𝑒𝑙𝑑% =
11.9 𝑔 𝑀𝑔𝑂
= 70.8%
16.8 𝑔 𝑀𝑔𝑂
The limiting reactant = Mg
Theoretical yield = 16.8 g MgO
Percent yield for the reaction = 70.8%
8.5
A KNO3 solution is made using 88.4 g of KNO3 and diluting to a total solution volume of 1.50 L.
Calculate the molarity and mass percent of the solution. (assume a density of 1.05 g/mL for the
solution.)
(5)
No of Moles of KNO3 = 88.4 g KNO3/101.11 g/mol= 0.87 mol KNO3
Molarity = No of moles/vol of solution
Molarity = 0.87 mol/1.50 L = 0.58 M
Mass of solution = density x vol = 1.05 g/mL x 1.50 x 103 mL) = 1.57 x 103 g
Mass% = Mass solute/mass of solution
Mass% = 88.4 g/(1.57 x 103 g) x 100% = 5.6%
8.6
Describe how you would make 2.5 L of a 0.1.. M KCl solution from a 5.5 M stock KCl solution.
(3)
Given M1 = 5.5 M
Find V1
M2 = 0.1 M
V2 = 2.5 L
M1V1=M2V2
𝑉1 =
𝑀2𝑉2
(0.1 𝑀)(2.5 𝐿)
=
= 0.045 𝐿
𝑀1
5.5 𝑀
Take 0.045 L of the stock in a 2.5 L volumetric flask and add 2.45 L of deionized water.
Multiple Choice
[25 Marks]
1) Which of the following statements about matter is FALSE?
(1)
A) Matter occupies space and has mass.
B) Matter exists in either a solid, liquid or gas state.
C) Matter is ultimately composed of atoms.
D) Matter is smooth and continuous.
E) none of the above
Answer: D
2) Which of the following is a homogeneous mixture?
A) trail mix
B) stainless steel
C) water
D) molten iron
E) none of the above
Answer: B
(1)
3) What is the correct formula for a potassium ion with 18 electrons?
A) P+
B) K+
C) KD) PE) none of the above
Answer: B
(1)
4) An atom that has the same number of neutrons as
(1)
A)
Cs
B)
Ba
C)
La
D)
Xe
E) none of the above
Answer: D
Ba is:
5) A student wrote the formula for the compound aluminum phosphate as AlPO4 . What is
wrong with this formula?
(1)
A) The compound is not charge-neutral.
B) Aluminum is a nonmetal so it cannot form an ionic compound.
C) The formula should be Al(PO4) .
D) There cannot be three different atom types in a chemical formula.
E) Nothing is wrong with the formula.
Answer: E
6) If 3.011 ×
molecules have a mass of 20.04 grams, what is the molar mass of this
substance?
(2)
A) 40.08 g/mol
B) 10.02 g/mol
C) 20.04 g/mol
D) 6.658 ×
g/mol
E) none of the above
Answer: A
7) If a sample of carbon dioxide contains 3.8 moles of oxygen atoms, how many moles of carbon
dioxide are in the sample?
(2)
A) 1.9
B) 3.8
C) 7.6
D) 11.4
E) none of the above
Answer: A
Diff: 3 Page Ref: 6.5
8) Identify the double displacement reactions among the following:
(1)
1. KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq)
2. Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)
3. H2SO4((aq) + 2NaOH(aq) → Na2SO4((aq) + 2H2O(l)
A) 1 and 2 only
B) 1 and 3 only
C) 2 and 3 only
D) All of 1, 2, and 3
E) None of 1, 2, and 3
Answer: D
9) The theoretical yield of a reaction is 75.0 grams of product and the actual yield is 42.0g. What
is the percent yield?
(3)
A) 75.0
B) 56.0
C) 31.5
D) 178
E) none of the above
Answer: B
10) What is the excess reactant for the following reaction given we have 3.4 moles of Ca(NO3)2
and 2.4 moles of Li3PO4?
Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2
(3)
A) Ca(NO3)2
B) Li3PO4
C) LiNO3
D) Ca3(PO4)2
E) not enough information
Answer: B
11) A 90.0 g sample of NaOH is dissolved in water and the solution is diluted to give a final
volume of 3.00 liters. The molarity of the final solution is ________.
(2)
A) 0.500 M
B) 0.750 M
C) 1.00 M
D) 2.25 M
E) none of the above
Answer: B
12) What is the concentration of sodium chloride in the final solution if
of
completely reacts and the total volume of the reaction is
, given the reaction:
(3)
Ba
(aq) + N S
(aq) → BaS
A) 0.226
B) 0.0259
C) 0.0519
D) 0.667
E) none of the above
Answer: C
(s) + 2NaCl (aq)
BaC
13) In the following reaction: NH4+ (aq) + H2O (aq) → NH3 (aq) + H3O+ (aq)
A) NH4+ is an acid and H2O is its conjugate base.
B) H2O is a base and NH3 is its conjugate acid.
C) NH4+ is an acid and H3O+ is its conjugate base.
D) H2O is a base and H3O+ is its conjugate acid.
E) NH4+ is a base and H2O is its conjugate acid.
Answer: D
14) A 25.0 mL sample of 0.105 M HCl was titrated with
concentration of the NaOH?
A) 0.0833 M
B) 0.132 M
C) 0.105 M
D) 0.075 M
E) none of the above
Answer: A
(1)
of NaOH. What is the
QUESTION 1
(3)
[9 MARKS]
1.1 How many significant figures are there in the following measurements?
(a) 0.770090 m 6 sig figs
(1)
(b) 4.1003 × 104 cm
(1)
5 sig figs
1.2 Perform the following operations and express the answers with the appropriate number of
significant figures.
(a) 320.5 – (6104.5/2.3)
(2)
= 320.5 – 2654.1304 
= - 2.3 x 103 
(b) [258.3 x 105) – (1.200 x 103)] x 2.8954
(2)
= 25828800 x 2.8954 
= 7.478 x 107 
1.3 Pure water freezes at 32 °F. If 10% by mass NaCl is added to the water, the freezing point
drops to about 20.3 °F. What is this difference expressed in °C
Difference = 32 – 20.3 = 11.7 °F 
°C = (°F – 32) x 5/9 
= (11.7 – 32) x 5/9
= - 11.3 C ° 
(3)
QUESTION 2
[11 MARKS]
A 33.0-g sample of an unknown liquid at 20.0 °C is heated to120 °C. During this heating, the
density of the liquid changes from 0.854 g/cm3 to 0.797 g/cm3. What volume would this sample
occupy at 120 °C?
(3)
V =
m
33.0 g
 =
= 41.405 = 41.4 cm3 
0.797 g/cm3
d
1.3 What is the volume in cubic millimeters of a 3.54 carat diamond, given that the density of
diamond is 3.51 g/mL? (1 carat = 200 mg). Assuming that the diamond is generally spherical
(volume of a sphere is V =
4 3
pr , what is the approximate diameter of the diamond in mm?
3
(8)
Vol of diamond = 3.54 carat (200/1carat)  (1g/1000mg)  (1cm3/3.51 g) 
= 0.201 cm3 
V = 4/3πr3 
0.197 = 4/3 πr3 
r = 0.363 cm 
2r = 0.7267 mm x 10mm/1cm = 7.27 mm 
QUESTION 3
3.1
3.2
3.3
[6]
Name the following compounds:
(a)
(NH4)2CO3
(b)
NiB
(Ammonium carbonate)
(1)
(Nickel(III) boride)
(1)
Give the molecular formula of each of the following compounds:
(a)
Lutetium(III) sulfate octahydrate
(b)
Tungsten(II) carbide
(Lu2(SO4)3•8H2O)
(1)
(W2C)
(1)
How many thallium cations are present in a 10.0 g sample of thallium(III) nitrate? Show all your
calculations.
Tl(NO3)3 = 328.42 g/mol
10.0 g/328.42 g/mol = 0.030448815 moles
1 mole Tl(NO3)3 contains 6.022 × 1023 Tl3+ cations
(2)
Therefore, 0.030448815 moles contain 1.834 × 1023 Tl3+ cations.
QUESTION 4
[5]
Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass, and has a molar
mass of 206 g/mol. Determine the molecular formula for ibuprofen:
(5)
QUESTION 5
[9]
In nature, copper occurs in a variety of minerals as sulfides, sulfosalts, carbonates and oxides. Examples
of copper containing minerals are: chalcopyrite [CuFeS2], enargite [Cu3AsS4], and malachite
[Cu2CO3(OH)2].
(a)
Which of the three ores above contains more copper by mass? Please show all your calculations.
(4.5)
(b)
Suppose a certain rock contains 0.75% CuO. What mass of this rock, in tons, must be processed in
order to obtain 750 kilograms of pure copper?
(4.5)
a) Mass% of Cu in chalcopyrite [CuFeS2]:
63.55/(63.55 + 55.85 + 32.07×2) × 100% = 34.62%
Mass% of Cu in enargite [Cu3AsS4]:
(3×63.55)/(3×63.55 + 74.92 + 32.07×4) × 100% = 48.41%
Mass% of Cu in malachite Cu2CO3(OH)2]:
(2×63.55)/(2×63.55 + 12.01 + 1.008×2 + 16.00×5) × 100% = 57.48%
b) The rock contains 0.75% CuO.
The mass percent of copper in CuO = 63.55/(63.55 + 16.00) × 100% = 79.8868636%
The mass percent of copper in the rock is thus 0.75 × 0.798868636 = 0.599%
The mass of rock to be processed = 750 kg/0.599 = 12520.87 kg = 12.52 tons.
QUESTION 6
Calculate the number of moles of KOH in 5.50 mL of a 0.360 M KOH solution. What is the pOH of the
solution?
Multiple Choice
(5)
[24]
1) A student attempts to balance a chemical equation and comes up with the following result:
8KClO3 + C12H22O10  8KCl + 12CO2 + 11H2O
Turns out he wrote the initial equation wrong so he could not balance the equation. Which
element(s) are not balanced in this result?
(1)
a. Cl
b. O
c. H
d. O and H
ANS: B
2) How many molecules of sulfur trioxide are in 78.0 grams?
b. 7.33 × 1023
a. 5.87 × 1023
d. 0.974
c. 3.76 × 1027
e. none of the above
(2)
ANS: A
3) A compound contains arsenic and oxygen as its only elements. A 1.626 g sample of the
compound contains 1.060 g of arsenic. What is the empirical formula for the compound?
(3)
a. As2O
b. AsO
c. As2O3
d. AsO2
e. As2O5
ANS E
4) An iron ore sample is found to be 35.00% Fe by mass. How many grams of ore are needed to
obtain
of Fe?
(3)
a. 1297
b. 158.9
c. 295.1
d. 350.0
e. none of the above
Answer: A
5) Which compound produces three ions per formula unit by dissociation when dissolved in
water?
(2)
a. sodium nitrate
c. calcium perchlorate
b. nickel sulfate
d. aluminum sulfate
e. ammonium bromate
ANS: C
6) Select the examples in which the formulas do not correctly match the names of the compounds
indicated.
(2)
I.
Sodium thiosulfate
Na2SO3
II. Barium oxalate
BaC2O4
III. Iron(II) sulfate hexahydrate
FeSO4·6H2O
IV. Calcium phosphate
Ca3PO4
a. II only
b. II and III
c. I, II and IV
d. I and IV
e. II, III and IV
ANS: D
7) What is the theoretical yield of a reaction if 25.0 grams of product were actually produced
from a reaction that has a 88% yield?
(3)
a. 28.4
b. 22.0
c. 3.52
d. 352
e. none of the above
Answer: A
8) The solubility of solids in water:
a. is independent of the temperature.
b. increases with increasing temperature.
c. decreases with increasing temperature.
d. Solids are not soluble in water.
(1)
e. none of the above
Answer: B
9) How many grams of a 23.4% by mass NaF solution is needed if you want to have
NaF?
(3)
a. 55.9
b. 31.1
c. 13.1
d. 239
e. none of the above
Answer: D
1) Which statement below is true?
of
(1)
a.
All acids are strong electrolytes and ionize completely when dissolved in water.
b.
All bases are weak electrolytes and ionize completely when dissolved in water.
c.
All bases are strong electrolytes and ionize completely when dissolved in water.
d. All salts are strong electrolytes and dissociate completely if they dissolve in water.
e. All salts are weak electrolytes and ionize partially when dissolved in water.
11) A 35.0 mL sample of 0.225 M HBr was titrated with 42.3 mL of KOH. What is the
concentration of the KOH?
(3)
a. 0.157 M
c. 0.272 M
e. none of the above
b. 0.303 M
d. 0.186 M
Answer: D
ANS: D
Long Questions
[45 MARKS]
Question 1
[10 MARKS]
1) Perform the following calculations and give the answer to the correct number of significant
figures.
a.
(0.311 x 0.57) ÷ 5.93
(2)
= 0.17727  ÷ 5.93
= 0.02989
= 0.030
2) Name or the give the symbol for the following
(3)
a) CoCl3 Cobalt(III)Chloride 
b) Tin(IV)sulphide SnS2 
c) KIO3 Potassium Iodate 
3) One litre of gasoline in an automobile’s engine produces on the average 9.5 kg of carbon dioxide,
which is a greenhouse gas, that is, it promotes the warming of Earth’s atmosphere. Calculate the
annual production of carbon dioxide in kilograms and µg if there are 40 million cars in the South
Africa and each car covers a distance of 5000 km at a consumption rate of 20 km per litre.
(5)
40 x 106 x (5000 km/car)  x (1 litre/20 km)  x (9.5 kg/1L) 
= 9.5 x 1010 kg CO2
In μg
= 9.5 x 1010 kg x (103 g/ 1kg)  x (106 ug/1 g)  = 9.5 x 1019 μg 
Question 2
[18 Marks]
1) When 20.0 g of A was heated, 8.8 g of B was given off, leaving 10.2 g of E. The same quantity of
B can also be prepared by a combination of 2.4 g of C and 3.2 g of D. The E can be electrolysed,
after melting, to yield 8.0 g of F and 1.6 g of G, neither of which can be further decomposed by
ordinary chemical means. Identify the lettered materials A, B, E, F and G as either an element or
a compound and explain your decision.
(6)
A + E can be decomposed   compound 
B is a combination of C and D   compound 
F + G can be broken down   elements 
2) At least 25 µg of tetrahydrocannabinol (THC), the active ingredient in marijuana, is required to
produce intoxication. The molecular formula of THC is C21H30O2. How many moles of THC does this
25 µg represent? How many molecules?
(5)
3) Which contains more oxygen by mass;
0.5 mol of potassium perchlorate;
55.3 g sodium hypochlorite;
59.8 g of diphosphorous pentoxide
(7)
Question 3
[18 MARKS]
1) You are given a range of solids to perform precipitation reactions
(5)
a) Predict the products of the precipitation reactions that occur when these solids are
added to water.
b) Provide the net ionic equations.
1)
Sodium chloride is added to lead(II)sulphate
2)
Sodium chloride is added to ammonium nitrate
3)
Sodium chloride is added to silver sulfate
2) Consider the reaction between calcium oxide and carbon dioxide
CaO(s) + CO2(g)
(6)
CaCO3(s)
Bongani allows 14.4 g of CaO and 13.8 g of CO2 to react. When the reaction is finished, he
collects 19.4 g of CaCO3. Find
i. the limiting reactant, CaO
ii. theoretical yield, 25.7 g CaCO3
iii. percent yield. 75.5%
14.4 𝑔 𝐶𝑎𝑂 𝑥
1 𝑚𝑜𝑙 𝐶𝑎𝑂
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3 100.09 𝑔 𝐶𝑎𝐶𝑂3
𝑥
𝑥
= 25.7 𝑔 𝐶𝑎𝐶𝑂3
56.08 𝑔 𝐶𝑎𝑂
1 𝑚𝑜𝑙 𝐶𝑎𝑂
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3
13.8 𝑔 𝐶𝑂2 𝑥
1 𝑚𝑜𝑙 𝐶𝑂2
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3 100.09 𝑔 𝐶𝑎𝐶𝑂3
𝑥
𝑥
= 31.4 𝑔 𝐶𝑎𝐶𝑂3
44.01 𝑔 𝐶𝑂2
1 𝑚𝑜𝑙 𝐶𝑂2
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3
𝑦𝑖𝑒𝑙𝑑% =
19.4 𝑔 𝐶𝑎𝐶𝑂3
𝑥 100% = 75.5%
25.7 𝑔 𝐶𝑎𝐶𝑂3
Question 4
1) Identify the acid-base conjugate pairs in the following reactions:
v)
HCO3-(aq) + H3O+(aq)
CO2(g) + H2O(l) + H2O(l)
(2)
HCO3- (base) and CO2 + H2O (acid),
H3O+ (acid) and H2O (base)
HSO4-(aq) + NO2+(aq) + H2O(l)
vi) H2SO4(aq) + HNO3(aq)
(2)
H2SO4 (acid) and HSO4- (base),
HNO3 (base) and NO2+ + H2O (acid)
2) A laboratory procedure calls for making 500.0 mL of a 1.4 M KNO3 solution. How much KNO3 in
grams is needed?
𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
(3)
𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠
𝑉𝑜𝑙.𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
𝑁𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑥 𝑉𝑜𝑙. 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
No. of moles of KNO3 = 1.4 mol/L x 500.0 mL = 700 mMol = 0.700 mol
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑚𝑎𝑠𝑠
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
Actual mass = No of Moles x Molar mass = 0.700 mol KNO3 x 101.11 g/mol = 70.78 g KNO3
QUESTION 1
(Multiple Choice)
[12]
(i) In a chemical reaction, mass is conserved. This means that
a) The mass of the reactants stays the same during a chemical reaction
b) The mass of the products stays the same during a chemical reaction
c) The type and number of atoms in the reactants equals the type and number of
atoms
in the products
d) The mass of the products is always twice the mass of the reactants.
Answer: c
(ii) In any sample of water, there are always some water molecules which have become ions.
These ions are
a) H2O+ and OHb) HO+ and H2Oc) H3O+ and OHd) HO+
and HO
Answer: c
(iii) If a solution is acidic, it can be neutralized by adding
a) A stronger acid
b) Heat
c) A base
Answer: c
d) A weaker acid
(iv) When carbon dioxide (CO2) gas reacts with water
a) A strong base is produced
b) The carbon and oxygen atoms disappear
c) An acid is produced
d) More carbon and oxygen atoms are produced
Answer: c
(v) What is the general form for a single-replacement reaction?
a) AX + BY  AY + BX
c) A + B  AB
b) AB  A + B
d) A +BX  AX + B
Answer: d
(vi) Which of the following structures represents the conjugate acid of HPO42- ?
a) H2PO4-
b) H3PO4
c) H4PO4+
d) PO43-
Answer: a
(vii) Which of the following relationships is true for an acidic solution at 25ºC?
a) [H+] > [OH-]
b) pH > 7.00
c) Kw > 1 x 10-14
d) The solution is negatively-charged
Answer: a
(viii) Which one of the following relationships is true in water at 25ºC?
a) [H+] = [H2O+]
b) [OH-] = [H2O-]
c) Kw > 1 x 10-14
d) [H+] = [OH-]
Answer: d
(ix) Which of the following is not an example of a weak acid?
a) Lactic acid
b) Carbonic acid
c) Sulfuric acid
d) Pyruvic acid
Answer: c
(x) Which of the following statements concerning a saturated solution is incorrect?
a) Undissolved solute must be present.
b) Undissolved solute may or may not be present.
c) Undissolved solute, if present, is continually dissolving.
d) Undissolved solute, if present, is in equilibrium with dissolved solute.
Answer: a
(xi) A crystal of solid NaCl is placed into an aqueous NaCl solution. It is observed that most, but not all,
of the crystal dissolves. This means the original solution was
a) supersaturated
b) unsaturated
d) more than one correct response
c) dilute
e) no correct response
Answer: b
(xii) A weak electrolyte exists predominantly as __________ in solution.
a) atoms
b) ions c) molecules
d) electrons e) an isotope
Answer: c
QUESTION 2
[13]
2.1
Perform the following calculations and give the answer to the correct number of significant figures.
(0.871 x 0.57) ÷ 5.971
(2)
= 0.49647 + 5.971
= 6.467
= 6.47
2.2
A submicroscopic particle suspended in a solution has a volume of 1.3 µm3. Convert this volume to
liters.
(2)
= 1.3 x (1m/106 µm)3 x (10-1 dm/1m)3
= 1.3 x 10-21 dm3
1 dm3 = 1 L
Therefore = 1.3 x 10-15L
2.3
Iodine is a bluish-black solid. It forms a violet-colored vapor when heated. The solid melts at 235°F.
What is this temperature in degrees Celsius and in Kelvin?
(2)
°C = (°F – 32) x 5/9
= (235 – 32) x 5/9
= 113 C °
K = °C + 273.15
K = 386.15
2.4
Write a complete ionic and net ionic equation, for the reaction shown below:
(3)
2 NH4Cl(aq) + Hg2(NO3)2(aq)  Hg2Cl2(s) + 2 NH4NO3(aq)
Answer: Complete ionic equation:
2 NH4+ (aq) + 2Cl- (aq) + Hg22+ (aq) + 2 NO3- (aq)  Hg2Cl2 (s) + 2NH4+ (aq) + 2 NO3- (aq)
Net ionic equation:
2 Cl- (aq) + Hg22+ (aq)  Hg2Cl2 (s)
2.5
For the reactions shown below(type of reaction is mentioned), write the complete and balanced
chemical reaction in each case.
a) SiI4 + Mg (single replacement)
Answer: SiI4 + 2Mg 2MgI2 + Si
b) Al + I2 (synthesis)
Answer: 2Al + 3I2  2AlI3
c) CuCl2 + KOH (double replacement)
Answer: CuCl2 + 2KOH  Cu(OH)2 + 2KCl
d) NH3 (decomposition)
Answer: 2NH3  N2 + 3H2
(4)
QUESTION 3
3.1
3.2
3.3
[6]
Name the following compounds:
(a)
VSO4•7H2O
(b)
SF6
Vanadium(II) sulfate heptahydrate
(1)
Sulfur hexafluoride
(1)
Give the molecular formula of each of the following compounds:
(a)
Tin(IV) chromate
Sn(CrO4)2
(1)
(b)
Thorium dicarbide
ThC2
(1)
How many rhodium cations are present in a 10.0 g sample of Rhodium(III) sulfate? Show all your
calculations.
(2)
Rhodium(III) sulfate = Rh2(SO4)3 = 494.01 g/mol
10.0 g/494.10 g/mol = 0.020242505 moles
1 mol of Rh2(SO4)3 contains 2×6.022×1023 Rh3+ cations
Therefore, 0.020242505 moles contain: 0.020242505×2×6.022×1023
= 2.44×1022 Rh3+ cations
QUESTION 4
[8]
Lead is found in Earth’s crust as several different lead ores mainly Galena (PbS), Cerussite (PbCO 3) and
Anglesite (PbSO4).
4.1
Which of the three ores above contains more lead by mass? Please show all your calculations.
(3)
%Mass of Pb in Galena (PbS): 207.2/(207.2 + 32.07) × 100% = 86.6%
%Mass of Pb in Cerussite (PbCO3): 207.2/[207.2 + 12.01 + (16.00×3)] × 100% = 77.5%
%Mass of Pb in Anglesite (PbSO4): 207.2/[207.2 + 32.07 + (16.00×4)] × 100% = 68.3%
4.2
Suppose a certain rock contains 38.0% PbS, 25.0% PbCO3 and 17.4% PbSO4, with the remainder of
the rock being substances that do not contain lead. How much of this rock, in kilograms, must be
processed to obtain 5.0 tons of lead?
(5)
Total Percentage by Mass of Lead in the rock from all three minerals:
PbS: (38%×86.6%) + PbCO3: 25%×77.5% + 17.4%×68.3% = 64.2%
This means 1000 g of this rock contains 642 g Pb or 1.0 kg contains 0.642 kg Pb
Thus, 5 tons = 5000 kg will be contained in:
5000/0.642 = 7788.2 kg = 7.79 tons of the rock.
QUESTION 5
5.1
[9]
Write equations which represent the dissociation of H3AsO4 (a polyprotic acid) in
aqueous solution. Show each step of dissociation.
(3)
Dissociation Step 1: H3AsO4(aq) + H2O(l) ⇌ H2AsO4-(aq) + H3O+(aq)
Dissociation Step 2: H2AsO4-(aq) + H2O(l) ⇌ HAsO42-(aq) + H3O+(aq)
Dissociation Step 3: HAsO42-(aq) + H2O(l) ⇌ AsO43-(aq) + H3O+(aq)
5.2
Write the acid-base reaction between CH3-CH2-COOH (acts as an acid) and
According to Brønsted–Lowry theory, identify each chemical as either an
identify “conjugate” relationships.
CH3MgBr.
‘acid’ or a ‘base’ and
(3)
CH3-CH2-COOH + CH3MgBr  CH3-CH2-COOMgBr + CH4
(Acid)
(Base)
(Base)
(Acid)
1ST Conjugate acid-base pair: CH3-CH2-COOH and CH3-CH2-COOMgBr
2ND Conjugate acid-base pair: CH4 and CH3MgBr .(1 mark)
5.3. What is the pOH of a 0.0235 M HCl solution?
(3)
Answers:
pH = -log [H+] = -log(0.0235) = 1.629
pOH = 14.000 – pH = 14.000 – 1.629 = 12.371
QUESTION 6
6.1
[14]
10.4 g of As reacts with 11.8 g of S to produce 14.2 g of As2S3. Find the limiting
reactant, theoretical yield and percent yield for this reaction. The balance chemical
equation is (show your calculations):
(6)
2 As (s) + 3 S (l)  As2S3 (s)
Answer: 10.4 𝑔 𝐴𝑠 𝑥
1 𝑚𝑜𝑙 𝐴𝑠
74.92 𝑔 𝐴𝑠
𝑥
1 𝑚𝑜𝑙 𝐴𝑠2𝑆3
2 𝑚𝑜𝑙 𝐴𝑠
𝑥
246.05 𝑔 𝐴𝑠2𝑆3
1 𝑚𝑜𝑙 𝐴𝑠2𝑆3
= 17.1 𝑔 𝐴𝑠2𝑆3 (least amount;
therefore As is limiting reagent
1 𝑚𝑜𝑙 𝑆
1 𝑚𝑜𝑙 𝐴𝑠2𝑆3
246.05 𝑔 𝐴𝑠2𝑆3
𝑥
𝑥
= 30.2 𝑔 𝐴𝑠2𝑆3
32.07 𝑔 𝑆
3 𝑚𝑜𝑙 𝑆
1 𝑚𝑜𝑙 𝐴𝑠2𝑆3
Theoretical yield is 17.1 g of As2S3.
11.8 𝑔 𝑆 𝑋
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑 =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑥 100%
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑 =
6.2
14.2 𝑔
17.1 𝑔
𝑥 100% = 83.0%
Describe how you would make 500.0 mL of a 0.200 M NaOH solution from a 15.0 M stock NaOH
solution.
(4)
Given M1 = 15.0 M
Find V1 and determine how much distilled water need to be added.
M2 = 0.200 M
V2 = 500.0 mL
From M1V1=M2V2
𝑉1 =
𝑀2𝑉2
(0.200 𝑀)(500.0 𝑚𝐿)
=
= 6.67 𝑚𝐿
𝑀1
15.0 𝑀
Therefore take 6.67 mL of the stock solution in a 500.0 mL volumetric flask and add 493.33 mL of
deionized water.
6.3
A laboratory procedure calls for making 500.0 mL of a 1.4 M KNO3 solution. How much KNO3 in
grams is needed?
𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
(4)
𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠
𝑉𝑜𝑙.𝑠𝑜𝑙𝑛 𝑖𝑛 𝐿
𝑁𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑥 𝑉𝑜𝑙. 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
No. of moles of KNO3 = 1.4 mol/L x 500.0 mL = 700 mMol = 0.700 mol
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑚𝑎𝑠𝑠
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
Actual mass = No of Moles x Molar mass = 0.700 mol KNO3 x 101.11 g/mol = 70.78 g KNO3
QUESTION 7
7.1.
[8]
A reaction of hydrogen and oxygen produces water. Calculate how much oxygen you need, in
grams, to react with 32.3 grams of hydrogen to produce water. (5)
Answer:
Balanced Chemical reaction: 2H2 + O2  2H2O.
Converting the mass of hydrogen to moles of hydrogen.
moles H 2 = 32.3 g H 2 ×
Calculating how many moles of oxygen are needed:
mole H 2
= 16.0 mole H 2
2.02 g H 2
moles O2 = 16.0 mole H 2 ×
1 mole O 2
= 8.00 mol O 2
2 mol H 2
Converting Moles of oxygen into grams of oxygen:
grams of O 2 =
7.2
32.00 grams O 2
× 8.00 mole O 2 = 256 grams of O 2
mole O 2
For the reaction:
(3)
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
a) Determine the number of moles H+ that are required to form 1.22 mol H2.
b) Determine the mass in grams of Zn that is required to form 0.621 mol of H2
Answer:
From the balanced equation, you can see that 2 mol H+ is used for every 1 mol H2.
If we use this as a conversion factor, then for 1.22 mol H2:
moles H+ = 1.22 mol H2 x 2 mol H+ / 1 mol H2
moles H+ = 2.44 mol H+
The atomic mass of zinc is 65.38, so there are 65.38 g in 1 mol Zn.
Plugging in these values gives us:
mass Zn = 0.621 mol H2 x 1 mol Zn / 1 mol H2 x 65.38 g Zn / 1 mol Zn
mass Zn = 40.6 g Zn