← Study Hub

IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 3: Stoichiometry -- Mock Exam 2
1 hour 15 minutes
80
7
75:00
0620

Instructions

Question Navigation

Question 1 -- Formulae from Ionic Charges and Valencies
Total: 12 marks
A student in Kolkata is learning to write chemical formulae using ionic charges and valencies.
(a) [4]
Use ionic charges to work out the formula of each compound.

(i) Potassium bromide (K+ and Br−)
(ii) Iron(III) chloride (Fe3+ and Cl−)
(iii) Zinc nitrate (Zn2+ and NO3−)
(iv) Calcium phosphate (Ca2+ and PO43−)
Model Answer -- 1(a)
Potassium bromide: KBr [1]
Iron(III) chloride: FeCl3 [1]
Zinc nitrate: Zn(NO3)2 [1]
Calcium phosphate: Ca3(PO4)2 [1]
(b) [3]
Copper(II) carbonate decomposes on heating.

(i) Write the word equation for this decomposition. [1]

(ii) Write a balanced symbol equation for this reaction. [1]

(iii) Add state symbols to your equation. [1]
Model Answer -- 1(b)
Copper(II) carbonate → copper(II) oxide + carbon dioxide [1]
CuCO3 → CuO + CO2 [1]
CuCO3(s) → CuO(s) + CO2(g) [1]
(c) [3]
Balance the following equations.

(i) ___ Al + ___ HCl → ___ AlCl3 + ___ H2 [1]

(ii) ___ Fe2O3 + ___ C → ___ Fe + ___ CO2 [1]

(iii) ___ C3H8 + ___ O2 → ___ CO2 + ___ H2O [1]
Model Answer -- 1(c)
2Al + 6HCl → 2AlCl3 + 3H2 [1]
2Fe2O3 + 3C → 4Fe + 3CO2 [1]
C3H8 + 5O2 → 3CO2 + 4H2O [1]
(d) [2]
(i) State the law of conservation of mass. [1]

(ii) Explain why a balanced equation must have equal numbers of each type of atom on both sides. [1]
Model Answer -- 1(d)
In a chemical reaction, matter is neither created nor destroyed / the total mass of reactants equals the total mass of products [1]
Because atoms are rearranged in a reaction but not created or destroyed, so each atom present in the reactants must appear in the products [1]
Question 2 -- Moles, Molar Mass and Avogadro Calculations
Total: 12 marks
A chemistry laboratory in Singapore has a set of samples for students to analyse.
(a) [4]
Calculate the number of moles in each of the following.

(i) 5.6 g of potassium hydroxide, KOH [1]

(ii) 9.8 g of sulfuric acid, H2SO4 [1]

(iii) 21.6 g of silver bromide, AgBr [2]
Model Answer -- 2(a)
Mr of KOH = 39 + 16 + 1 = 56; Moles = 5.6 / 56 = 0.1 mol [1]
Mr of H2SO4 = 2 + 32 + 64 = 98; Moles = 9.8 / 98 = 0.1 mol [1]
Mr of AgBr = 108 + 80 = 188 [1]
Moles of AgBr = 21.6 / 188 = 0.115 mol (3 s.f.) [1]
(b) [3]
(i) Calculate the number of molecules in 8.8 g of carbon dioxide, CO2. [2]

(ii) Calculate the mass of one molecule of water, H2O. Give your answer in standard form. [1]
Model Answer -- 2(b)
Moles of CO2 = 8.8 / 44 = 0.2 mol [1]
Molecules = 0.2 × 6.02 × 1023 = 1.204 × 1023 [1]
Mass of 1 molecule of H2O = 18 / (6.02 × 1023) = 2.99 × 10−23 g [1]
(c) [3]
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass.

(i) Determine the empirical formula of this hydrocarbon. [2]

(ii) The Mr of the hydrocarbon is 28. Determine its molecular formula. [1]
Model Answer -- 2(c)
C: 85.7/12 = 7.14   H: 14.3/1 = 14.3   Ratio: 7.14/7.14 = 1, 14.3/7.14 = 2 [1]
Empirical formula = CH2 [1]
Empirical formula mass = 14; n = 28/14 = 2; Molecular formula = C2H4 (ethene) [1]
(d) [2]
Calculate the percentage by mass of oxygen in copper(II) sulfate, CuSO4.
Model Answer -- 2(d)
Mr of CuSO4 = 64 + 32 + (4×16) = 160 [1]
% O = (64/160) × 100 = 40.0% [1]
Question 3 -- Reacting Masses and Stoichiometric Calculations
Total: 12 marks
A steel works in Sheffield uses the blast furnace process to extract iron from iron ore.
(a) [4]
Iron is extracted from haematite, Fe2O3, by reduction with carbon monoxide:

Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

Calculate the mass of iron that can be obtained from 800 g of iron(III) oxide.
Model Answer -- 3(a)
Mr of Fe2O3 = (2×56) + (3×16) = 160 [1]
Moles of Fe2O3 = 800 / 160 = 5.0 mol [1]
From equation: 1 mol Fe2O3 → 2 mol Fe; so moles of Fe = 5.0 × 2 = 10.0 mol [1]
Mass of Fe = 10.0 × 56 = 560 g [1]
(b) [4]
A student adds 6.5 g of zinc to a solution containing 7.3 g of hydrochloric acid (HCl).

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

(i) Calculate the moles of zinc and the moles of HCl. [2]

(ii) Determine which reagent is the limiting reagent. Explain your answer. [1]

(iii) Calculate the mass of hydrogen gas produced. [1]
Model Answer -- 3(b)
Moles of Zn = 6.5 / 65 = 0.1 mol [1]
Moles of HCl = 7.3 / 36.5 = 0.2 mol [1]
Ratio needed: 1 mol Zn : 2 mol HCl. We have 0.1 mol Zn and 0.2 mol HCl. Exact stoichiometric ratio -- neither is in excess (or both just used up). [1]
0.1 mol Zn → 0.1 mol H2; Mass = 0.1 × 2 = 0.2 g [1]
(c) [4]
Sodium carbonate reacts with hydrochloric acid:

Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

Calculate the mass of sodium carbonate needed to produce 5.85 g of sodium chloride.
Model Answer -- 3(c)
Mr of NaCl = 23 + 35.5 = 58.5; Moles NaCl = 5.85 / 58.5 = 0.1 mol [1]
From equation: 1 mol Na2CO3 → 2 mol NaCl; so moles Na2CO3 = 0.1 / 2 = 0.05 mol [1]
Mr of Na2CO3 = (2×23) + 12 + (3×16) = 106 [1]
Mass = 0.05 × 106 = 5.3 g [1]
Question 4 -- Percentage Yield and Percentage Purity
Total: 12 marks
A cement factory near Chennai processes limestone to produce quicklime (calcium oxide) and uses the product in construction.
(a) [4]
CaCO3(s) → CaO(s) + CO2(g)

250 g of calcium carbonate is heated strongly. The actual yield of calcium oxide is 119 g.

(i) Calculate the theoretical yield of calcium oxide. [2]

(ii) Calculate the percentage yield. [2]
Model Answer -- 4(a)
Moles CaCO3 = 250 / 100 = 2.5 mol [1]
Moles CaO = 2.5 mol; Theoretical mass = 2.5 × 56 = 140 g [1]
% yield = (actual / theoretical) × 100 [1]
% yield = (119 / 140) × 100 = 85.0% [1]
(b) [4]
A 10.0 g sample of impure calcium carbonate is reacted with excess hydrochloric acid. The volume of carbon dioxide collected at RTP is 1920 cm³.

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

Calculate the percentage purity of the calcium carbonate sample.
Model Answer -- 4(b)
Moles of CO2 = 1920 / 24000 = 0.08 mol [1]
From equation: 1 mol CaCO3 → 1 mol CO2; so moles CaCO3 = 0.08 mol [1]
Mass of pure CaCO3 = 0.08 × 100 = 8.0 g [1]
% purity = (8.0 / 10.0) × 100 = 80.0% [1]
(c) [2]
Give two reasons why the percentage yield in an industrial process is often less than 100%.
Model Answer -- 4(c)
Some product is lost during transfer between containers / purification / filtration [1]
The reaction may be reversible / incomplete, or side reactions may produce unwanted by-products [1]
(d) [2]
For the thermal decomposition of calcium carbonate:

CaCO3 → CaO + CO2

(i) State the formula for atom economy. [1]

(ii) Calculate the atom economy for producing CaO in this reaction. [1]
Model Answer -- 4(d)
Atom economy = (Mr of desired product / total Mr of all products) × 100 [1]
Atom economy = (56 / (56 + 44)) × 100 = (56/100) × 100 = 56.0% [1]
Question 5 -- Gas Volume Calculations at RTP
Total: 10 marks
In a school laboratory in Bristol, students investigate reactions that produce gases and measure their volumes at room temperature and pressure (RTP).
(a) [3]
Sodium reacts with water:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Calculate the volume of hydrogen gas produced at RTP when 2.3 g of sodium reacts with excess water.
Model Answer -- 5(a)
Moles of Na = 2.3 / 23 = 0.1 mol [1]
From equation: 2 mol Na → 1 mol H2; so moles H2 = 0.1 / 2 = 0.05 mol [1]
Volume = 0.05 × 24 = 1.2 dm³ (or 1200 cm³) [1]
(b) [3]
A student collects 4800 cm³ of carbon dioxide gas at RTP from the reaction of calcium carbonate with hydrochloric acid.

CaCO3 + 2HCl → CaCl2 + H2O + CO2

Calculate the mass of calcium carbonate that reacted.
Model Answer -- 5(b)
Moles of CO2 = 4800 / 24000 = 0.2 mol [1]
From equation: 1 mol CaCO3 → 1 mol CO2; so moles CaCO3 = 0.2 mol [1]
Mass = 0.2 × 100 = 20.0 g [1]
(c) [2]
In the combustion of methane:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

(i) State the ratio of volumes of CH4 : O2 : CO2 (at the same temperature and pressure). [1]

(ii) Calculate the volume of oxygen needed to completely burn 120 cm³ of methane. [1]
Model Answer -- 5(c)
Volume ratio CH4 : O2 : CO2 = 1 : 2 : 1 [1]
Volume of O2 = 120 × 2 = 240 cm³ [1]
(d) [2]
(i) Explain why one mole of any gas occupies the same volume at the same temperature and pressure. [1]

(ii) State the conditions for RTP (room temperature and pressure). [1]
Model Answer -- 5(d)
At the same temperature and pressure, equal numbers of moles of gas contain the same number of molecules, and the volume of a gas depends on the number of molecules (not their size) since gas molecules are far apart [1]
RTP conditions: approximately 20 °C (293 K) and 1 atmosphere (101 kPa) [1]
Question 6 -- Titration and Concentration Calculations
Total: 12 marks
A student in Bangalore carries out a titration to find the concentration of a hydrochloric acid solution.
(a) [3]
The student dissolves 10.6 g of anhydrous sodium carbonate (Na2CO3) in water and makes the volume up to 1.00 dm³.

(i) Calculate the concentration of the sodium carbonate solution in mol/dm³. [2]

(ii) Calculate the concentration in g/dm³. [1]
Model Answer -- 6(a)
Mr of Na2CO3 = (2×23) + 12 + (3×16) = 106; Moles = 10.6 / 106 = 0.1 mol [1]
Concentration = 0.1 / 1.00 = 0.1 mol/dm³ [1]
Concentration in g/dm³ = 10.6 / 1.00 = 10.6 g/dm³ [1]
(b) [5]
25.0 cm³ of the sodium carbonate solution is titrated with hydrochloric acid:

Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

The mean titre is 25.0 cm³ of HCl.

(i) Calculate the moles of sodium carbonate in 25.0 cm³ of solution. [1]

(ii) Use the mole ratio to find the moles of HCl. [1]

(iii) Calculate the concentration of HCl in mol/dm³. [1]

(iv) Calculate the concentration of HCl in g/dm³. [2]
Model Answer -- 6(b)
Moles Na2CO3 = 0.1 × (25.0/1000) = 0.0025 mol [1]
From equation: 1 mol Na2CO3 reacts with 2 mol HCl; moles HCl = 0.0025 × 2 = 0.005 mol [1]
Concentration HCl = 0.005 / 0.025 = 0.2 mol/dm³ [1]
Mr of HCl = 1 + 35.5 = 36.5 [1]
Concentration = 0.2 × 36.5 = 7.3 g/dm³ [1]
(c) [4]
A student needs to prepare 500 cm³ of 0.05 mol/dm³ NaOH from a stock solution of 0.5 mol/dm³ NaOH.

(i) Calculate the number of moles of NaOH needed in the diluted solution. [1]

(ii) Calculate the volume of stock solution required. [2]

(iii) Describe briefly how the student would prepare the diluted solution. [1]
Model Answer -- 6(c)
Moles needed = 0.05 × 0.5 = 0.025 mol [1]
Volume of stock = moles / concentration = 0.025 / 0.5 = 0.05 dm³ [1]
= 50 cm³ [1]
Measure 50 cm³ of the stock solution using a measuring cylinder or pipette, transfer to a 500 cm³ volumetric flask, and add distilled water up to the 500 cm³ mark [1]
Question 7 -- Water of Crystallisation and Mixed Calculations
Total: 10 marks
A student in Manchester heats hydrated copper(II) sulfate crystals to determine the number of water molecules in the formula.
(a) [2]
(i) Define the term water of crystallisation. [1]

(ii) State the difference between a hydrated salt and an anhydrous salt. [1]
Model Answer -- 7(a)
Water of crystallisation is water molecules that are chemically bonded / incorporated into the crystal structure of a salt [1]
A hydrated salt contains water of crystallisation; an anhydrous salt has had the water removed / contains no water of crystallisation [1]
(b) [5]
A student heats 6.25 g of hydrated copper(II) sulfate (CuSO4·nH2O). After heating until constant mass, the residue has a mass of 4.00 g.

Determine the value of n.
Model Answer -- 7(b)
Mass of water lost = 6.25 − 4.00 = 2.25 g [1]
Moles of CuSO4 = 4.00 / 160 = 0.025 mol [1]
Moles of H2O = 2.25 / 18 = 0.125 mol [1]
Ratio: CuSO4 : H2O = 0.025 : 0.125 [1]
Divide by smallest: 1 : 5, so n = 5 -- the formula is CuSO4·5H2O [1]
(c) [3]
(i) Calculate the percentage by mass of water in CuSO4·5H2O. [2]

(ii) State one observation the student would make during the heating. [1]
Model Answer -- 7(c)
Mr of CuSO4·5H2O = 160 + (5×18) = 250 [1]
% water = (90/250) × 100 = 36.0% [1]
The blue crystals turn to a white powder (as the hydrated salt loses its water of crystallisation) [1]

Self-Assessment

Tick marks earned, then click Calculate Grade.

0
80
0%
IGCSE Chemistry 0620 -- Paper 4 Theory (Extended) -- Topic 3: Stoichiometry -- Mock Exam 2
For examination preparation only. Not an official Cambridge Assessment paper.