IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 3: Stoichiometry -- Mock Exam 1

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
A_r values: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, Al = 27, S = 32, Cl = 35.5, K = 39, Ca = 40, Fe = 56, Cu = 64, Zn = 65, Br = 80, Ag = 108
Molar volume of gas at RTP = 24 dm³/mol
Avogadro constant = 6.02 × 10²³ /mol
Your answers will be automatically graded when you submit.

Question Navigation

Question 1 -- Chemical Formulae and Balancing Equations

Total: 12 marks

A chemistry teacher in Mumbai demonstrates several reactions to her class and asks them to write balanced equations.

(a) [4]

Write the chemical formula for each of the following compounds.

(i) Magnesium oxide
(ii) Calcium hydroxide
(iii) Aluminium sulfate
(iv) Ammonium nitrate

Model Answer -- 1(a)

Magnesium oxide: MgO [1]

Calcium hydroxide: Ca(OH)₂ [1]

Aluminium sulfate: Al₂(SO₄)₃ [1]

Ammonium nitrate: NH₄NO₃ [1]

⚠ If you missed marks here: Brackets are needed whenever a group ion is multiplied: Ca(OH)₂, not CaOH₂ (which would mean one O and two H), and Al₂(SO₄)₃, where 2 × Al³⁺ (+6) balances 3 × SO₄²⁻ (−6). Al₃(SO₄)₂ has the numbers the wrong way round, and its charges no longer cancel.

1 mark -- MgO correctly written 1 mark -- Ca(OH)2 correctly written 1 mark -- Al2(SO4)3 correctly written 1 mark -- NH4NO3 correctly written

(b) [4]

Balance the following equations.

(i) ___ Fe + ___ O₂ → ___ Fe₂O₃ [1]

(ii) ___ Na + ___ H₂O → ___ NaOH + ___ H₂ [1]

(iii) ___ CaCO₃ + ___ HCl → ___ CaCl₂ + ___ H₂O + ___ CO₂ [1]

(iv) ___ C₂H₆ + ___ O₂ → ___ CO₂ + ___ H₂O [1]

Model Answer -- 1(b)

4Fe + 3O₂ → 2Fe₂O₃ [1]

2Na + 2H₂O → 2NaOH + H₂ [1]

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ [1]

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O [1]

⚠ If you missed marks here: Balance only by changing the numbers in FRONT of formulae – altering a formula (writing H instead of H₂, or FeO instead of Fe₂O₃) scores nothing. For ethane the oxygen first comes out as 3½ O₂, so double everything: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.

1 mark -- 4Fe 3O2 2Fe2O3 balanced 1 mark -- 2Na 2H2O 2NaOH H2 balanced 1 mark -- CaCO3 2HCl CaCl2 H2O CO2 balanced 1 mark -- 2C2H6 7O2 4CO2 6H2O balanced

(i) State what each of the following state symbols represents: (s), (l), (g), (aq). [1]

(ii) Write a balanced equation with state symbols for the reaction of hydrochloric acid with sodium hydroxide solution. [1]

Model Answer -- 1(c)

(s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous / dissolved in water [1]

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) [1]

⚠ If you missed marks here: Water made in a neutralisation is a liquid, so it is H₂O(l), not H₂O(aq) – this is the commonest slip. The sodium chloride stays dissolved, so it is NaCl(aq), not NaCl(s); every formula needs its state symbol for the mark.

1 mark -- solid liquid gas aqueous dissolved water 1 mark -- HCl NaOH NaCl H2O correct state symbols balanced

(d) [2]

Consider the equation: 2Mg(s) + O₂(g) → 2MgO(s)

(i) State the ratio of moles of magnesium to moles of oxygen in this reaction. [1]

(ii) Explain, in terms of atoms and molecules, what this equation tells us. [1]

Model Answer -- 1(d)

The mole ratio of magnesium to oxygen is 2 : 1 [1]

Two atoms of magnesium react with one molecule of oxygen to form two formula units of magnesium oxide [1]

⚠ If you missed marks here: Read the ratio from the numbers in FRONT of the formulae, in the order asked: Mg : O₂ = 2 : 1 (writing 1 : 2 reverses it). The equation does not mean 2 g of magnesium reacts with 1 g of oxygen – the numbers count atoms, molecules or moles, not grams.

1 mark -- mole ratio 2 to 1 magnesium oxygen 1 mark -- two atoms magnesium react molecule oxygen form two magnesium oxide

Question 2 -- Relative Atomic Mass and Relative Molecular Mass

Total: 12 marks

A student at a school in London is studying relative masses and how they are used in chemical calculations.

(a) [3]

(i) Define relative atomic mass (A_r). [2]

(ii) State what relative molecular mass (M_r) means. [1]

Model Answer -- 2(a)

Relative atomic mass is the weighted average mass of naturally occurring atoms of an element [1]

on a scale where one atom of carbon-12 has a mass of exactly 12 [1]

Relative molecular mass is the sum of all the relative atomic masses of the atoms in a molecule [1]

⚠ If you missed marks here: Relative atomic mass needs both halves: a weighted AVERAGE of the isotopes AND the carbon-12 scale – "the mass of an atom" or "the number of protons and neutrons" misses both. It has no units because it is relative, so giving it in grams loses the mark; M_r is simply the sum of the A_r values of all the atoms in the formula.

1 mark -- weighted average mass naturally occurring atoms element 1 mark -- scale carbon-12 mass exactly twelve 1 mark -- sum relative atomic masses atoms molecule

(b) [4]

Calculate the relative molecular mass (M_r) of each of the following compounds.

(i) Sulfuric acid, H₂SO₄ [1]

(ii) Calcium carbonate, CaCO₃ [1]

(iii) Copper(II) sulfate pentahydrate, CuSO₄·5H₂O [1]

(iv) Aluminium hydroxide, Al(OH)₃ [1]

Model Answer -- 2(b)

H₂SO₄: (2×1) + 32 + (4×16) = 98 [1]

CaCO₃: 40 + 12 + (3×16) = 100 [1]

CuSO₄·5H₂O: 64 + 32 + (4×16) + 5×(2×1 + 16) = 250 [1]

Al(OH)₃: 27 + 3×(16 + 1) = 78 [1]

⚠ If you missed marks here: A number outside a bracket multiplies EVERYTHING inside it: Al(OH)₃ = 27 + 3 × (16 + 1) = 78, not 27 + 16 + 3 = 46. In CuSO₄·5H₂O the 5 multiplies the whole water molecule, adding 5 × 18 = 90 to 160; an answer of 178 means you counted only one water.

1 mark -- H2SO4 Mr equals 98 1 mark -- CaCO3 Mr equals 100 1 mark -- CuSO4 5H2O Mr equals 250 1 mark -- Al(OH)3 Mr equals 78

Calculate the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃.

Model Answer -- 2(c)

M_r of NH₄NO₃ = 14 + (4×1) + 14 + (3×16) = 80 [1]

Mass of N in formula = 2 × 14 = 28 [1]

% N = (28 / 80) × 100 = 35.0% [1]

⚠ If you missed marks here: Ammonium nitrate contains TWO nitrogen atoms (one in NH₄⁺ and one in NO₃⁻), so the mass of nitrogen is 2 × 14 = 28; using 14 gives 17.5%, exactly half the right answer. Divide by the M_r of the whole compound, 80, and multiply by 100 to get 35.0%.

1 mark -- Mr NH4NO3 equals 80 1 mark -- mass nitrogen 28 1 mark -- percentage nitrogen 35 percent

(d) [2]

Chlorine has two isotopes: ³⁵Cl (75.0%) and ³⁷Cl (25.0%).

Calculate the relative atomic mass of chlorine. Give your answer to one decimal place.

Model Answer -- 2(d)

A_r = (75.0 × 35 + 25.0 × 37) / 100 = (2625 + 925) / 100 [1]

A_r = 3550 / 100 = 35.5 [1]

⚠ If you missed marks here: The two isotopes are not equally common, so (35 + 37) ÷ 2 = 36 is wrong; three-quarters of chlorine atoms are chlorine-35, which pulls the weighted average down to 35.5. Multiply each mass by its percentage, add, then divide by 100.

1 mark -- correct formula working abundances multiplied 1 mark -- answer 35.5

Question 3 -- The Mole and Avogadro Constant

Total: 12 marks

A student in Delhi is learning about the mole concept and how to convert between mass, moles and number of particles.

(a) [2]

(i) State what is meant by the term mole. [1]

(ii) State the value of the Avogadro constant and its significance. [1]

Model Answer -- 3(a)

A mole is the amount of substance that contains 6.02 × 10²³ particles (atoms, molecules, ions or electrons) [1]

The Avogadro constant is 6.02 × 10²³ per mol; it is the number of particles in one mole of any substance [1]

⚠ If you missed marks here: A mole is an AMOUNT – a number of particles – not a mass, so "6.02 × 10²³ grams" or "the mass of one atom" loses the mark. Write the Avogadro constant in proper standard form, 6.02 × 10²³ per mole; "6.02 × 23" or a missing power of ten is a different number.

1 mark -- mole amount substance contains 6.02 times 10 power 23 particles 1 mark -- Avogadro constant 6.02 times 10 power 23 per mol number particles

(b) [4]

Calculate the number of moles in each of the following.

(i) 11.2 g of iron, Fe [1]

(ii) 4.4 g of carbon dioxide, CO₂ [1]

(iii) 0.78 g of aluminium oxide, Al₂O₃ [2]

Model Answer -- 3(b)

Moles of Fe = 11.2 / 56 = 0.2 mol [1]

M_r of CO₂ = 12 + (2×16) = 44; Moles = 4.4 / 44 = 0.1 mol [1]

M_r of Al₂O₃ = (2×27) + (3×16) = 102 [1]

Moles = 0.78 / 102 = 0.00765 mol (or 7.65 × 10⁻³ mol) [1]

⚠ If you missed marks here: Moles = mass ÷ M_r, never M_r ÷ mass – getting 131 mol from 0.78 g is a sign you divided the wrong way. For Al₂O₃, count every atom: (2 × 27) + (3 × 16) = 102, then 0.78 ÷ 102 = 0.00765 mol.

1 mark -- iron 0.2 mol 1 mark -- carbon dioxide 0.1 mol 1 mark -- Mr Al2O3 equals 102 1 mark -- moles Al2O3 0.00765 mol

Calculate the mass of 0.25 mol of calcium carbonate, CaCO₃.

Model Answer -- 3(c)

M_r of CaCO₃ = 40 + 12 + (3×16) = 100 [1]

Mass = moles × M_r = 0.25 × 100 = 25.0 g [1]

⚠ If you missed marks here: Mass = moles × M_r = 0.25 × 100 = 25.0 g; dividing instead (0.25 ÷ 100 = 0.0025 g) gives an impossibly small mass. Check the M_r first: CaCO₃ has THREE oxygens, so 40 + 12 + 48 = 100, not 68.

1 mark -- Mr CaCO3 equals 100 1 mark -- mass equals 25 grams

(d) [4]

(i) Calculate the number of molecules in 0.5 mol of water. [1]

(ii) Calculate the number of atoms in 0.5 mol of water. [1]

(iii) A sample contains 1.806 × 10²⁴ atoms of neon. Calculate the number of moles of neon in the sample. [2]

Model Answer -- 3(d)

Molecules = 0.5 × 6.02 × 10²³ = 3.01 × 10²³ molecules [1]

Each H₂O has 3 atoms, so total atoms = 3 × 3.01 × 10²³ = 9.03 × 10²³ atoms [1]

Moles = number of atoms / Avogadro constant [1]

Moles = 1.806 × 10²⁴ / 6.02 × 10²³ = 3.0 mol [1]

⚠ If you missed marks here: Each water molecule contains THREE atoms (2 H + 1 O), so the atom count is 3 × 3.01 × 10²³ = 9.03 × 10²³; reusing the molecule count, or multiplying by 2, loses (ii). For neon, moles = number of atoms ÷ (6.02 × 10²³) = 3.0 mol – multiplying instead gives a meaningless number near 10⁴⁸.

1 mark -- 3.01 times 10 power 23 molecules 1 mark -- 9.03 times 10 power 23 atoms three atoms per water 1 mark -- correct formula divide Avogadro constant 1 mark -- neon 3.0 mol

Question 4 -- Reacting Masses

Total: 12 marks

In a laboratory in Pune, students carry out experiments to investigate reacting masses in chemical reactions.

(a) [3]

Zinc reacts with hydrochloric acid as shown:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Calculate the mass of zinc chloride produced when 13.0 g of zinc reacts with excess hydrochloric acid.

Model Answer -- 4(a)

Moles of Zn = 13.0 / 65 = 0.2 mol [1]

From equation: 1 mol Zn produces 1 mol ZnCl₂, so moles of ZnCl₂ = 0.2 mol [1]

M_r of ZnCl₂ = 65 + (2×35.5) = 136; Mass = 0.2 × 136 = 27.2 g [1]

⚠ If you missed marks here: Zinc chloride is ZnCl₂, so its M_r is 65 + (2 × 35.5) = 136; using one chlorine (100.5) gives 20.1 g. The 2 in front of HCl does not carry over to the product: 1 Zn → 1 ZnCl₂, so 0.2 mol of zinc gives 0.2 mol, not 0.4 mol, of zinc chloride.

1 mark -- moles zinc 0.2 mol 1 mark -- mole ratio 1 to 1 moles ZnCl2 0.2 1 mark -- mass ZnCl2 27.2 grams

(b) [4]

Magnesium reacts with oxygen:

2Mg(s) + O₂(g) → 2MgO(s)

A student heats 4.8 g of magnesium in a crucible with 4.8 g of oxygen.

(i) Calculate the moles of magnesium and the moles of oxygen. [2]

(ii) Determine which reagent is in excess and calculate the mass of magnesium oxide produced. [2]

Model Answer -- 4(b)

Moles of Mg = 4.8 / 24 = 0.2 mol [1]

Moles of O₂ = 4.8 / 32 = 0.15 mol [1]

Ratio needed is 2 mol Mg : 1 mol O₂. For 0.2 mol Mg, we need 0.1 mol O₂. We have 0.15 mol O₂, so oxygen is in excess and magnesium is the limiting reagent. [1]

0.2 mol Mg → 0.2 mol MgO; Mass = 0.2 × 40 = 8.0 g [1]

⚠ If you missed marks here: Oxygen gas is O₂, so use M_r = 32: 4.8 ÷ 32 = 0.15 mol (dividing by 16 gives 0.3). Do not call oxygen limiting just because 0.15 < 0.2: the equation needs only HALF as much O₂ as Mg (0.1 mol), so magnesium runs out first and 8.0 g of MgO forms – 9.6 g (the two masses added) ignores the oxygen left over.

1 mark -- moles magnesium 0.2 mol 1 mark -- moles oxygen 0.15 mol 1 mark -- oxygen excess magnesium limiting reagent 1 mark -- mass MgO 8.0 grams

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass.

Determine the empirical formula of this compound.

Model Answer -- 4(c)

C: 40.0 / 12 = 3.33 H: 6.7 / 1 = 6.7 O: 53.3 / 16 = 3.33 [1] -- divide by A_r

Divide by smallest (3.33): C = 1, H = 2.01, O = 1 [1] -- ratio

Empirical formula = CH₂O [1]

⚠ If you missed marks here: Divide each percentage by its A_r (12, 1 and 16) – not by its proton number, and not by the other percentages – to get 3.33 : 6.7 : 3.33. Then divide by the smallest, 3.33, to get 1 : 2.01 : 1; round 2.01 to 2, giving CH₂O.

1 mark -- divide masses atomic masses correctly 1 mark -- divide smallest ratio correctly 1 mark -- empirical formula CH2O

(d) [2]

The compound in part (c) has a relative molecular mass of 180.

Determine its molecular formula.

Model Answer -- 4(d)

Empirical formula mass of CH₂O = 12 + 2 + 16 = 30; n = 180 / 30 = 6 [1]

Molecular formula = C₆H₁₂O₆ (glucose) [1]

⚠ If you missed marks here: n = 180 ÷ 30 = 6 multiplies EVERY subscript in CH₂O, giving C₆H₁₂O₆. "6CH₂O" means six separate formula units, and C₆H₂O₆ forgets to multiply the hydrogen.

1 mark -- empirical formula mass 30 ratio 6 1 mark -- molecular formula C6H12O6

Question 5 -- Percentage Yield and Percentage Purity

Total: 10 marks

A pharmaceutical company in Hyderabad manufactures calcium oxide by heating limestone (calcium carbonate) in a kiln.

(a) [4]

CaCO₃(s) → CaO(s) + CO₂(g)

A kiln is loaded with 500 g of calcium carbonate.

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

(ii) The actual yield of calcium oxide is 252 g. Calculate the percentage yield. [2]

Model Answer -- 5(a)

Moles of CaCO₃ = 500 / 100 = 5.0 mol [1]

1 mol CaCO₃ → 1 mol CaO; mass CaO = 5.0 × 56 = 280 g (theoretical) [1]

% yield = (actual / theoretical) × 100 [1]

% yield = (252 / 280) × 100 = 90.0% [1]

⚠ If you missed marks here: The theoretical yield is the mass of CaO that 500 g of CaCO₃ can make, 5.0 mol × 56 = 280 g – not 500 g, because 220 g leaves as CO₂. Percentage yield = actual ÷ theoretical × 100 = 252 ÷ 280 × 100 = 90.0%; an answer of 111% means the fraction is upside down.

1 mark -- moles CaCO3 5.0 1 mark -- theoretical yield CaO 280 grams 1 mark -- percentage yield formula correct 1 mark -- percentage yield 90 percent

(b) [2]

Suggest two reasons why the percentage yield of calcium oxide is less than 100%.

Model Answer -- 5(b)

The reaction may be incomplete / not all the CaCO₃ may have decomposed [1]

Some product may have been lost during transfer / mechanical losses / side reactions may have occurred [1]

⚠ If you missed marks here: Vague answers such as "human error" or "experimental error" score nothing – name what actually happened to the material. Two that earn marks: not all of the CaCO₃ decomposed (the heating was incomplete), and some product was lost, for example when the solid was taken out of the kiln.

1 mark -- reaction incomplete not all decomposed 1 mark -- product lost transfer mechanical losses side reactions

A sample of impure iron ore has a mass of 20.0 g. It contains iron(III) oxide, Fe₂O₃, and sand (silicon dioxide, SiO₂) as an impurity. When the sample is reacted with excess hydrochloric acid, only the Fe₂O₃ reacts. The mass of pure Fe₂O₃ in the sample is 16.0 g.

(i) Calculate the percentage purity of the iron ore sample. [1]

(ii) Calculate the mass of iron that could be obtained from the 16.0 g of Fe₂O₃. [3]

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Model Answer -- 5(c)

% purity = (16.0 / 20.0) × 100 = 80.0% [1]

M_r of Fe₂O₃ = (2×56) + (3×16) = 160; Moles = 16.0 / 160 = 0.1 mol [1]

From equation: 1 mol Fe₂O₃ → 2 mol Fe; so moles of Fe = 0.2 mol [1]

Mass of Fe = 0.2 × 56 = 11.2 g [1]

⚠ If you missed marks here: Purity = mass of pure substance ÷ mass of sample × 100 = 16.0 ÷ 20.0 × 100 = 80.0%; anything over 100% means the fraction is upside down. For the iron, use only the 16.0 g of Fe₂O₃ (the sand gives no iron) and remember each Fe₂O₃ gives TWO Fe: 0.1 mol → 0.2 mol → 11.2 g, so 5.6 g means the 2 was missed.

1 mark -- percentage purity 80 percent 1 mark -- moles Fe2O3 0.1 mol Mr 160 1 mark -- mole ratio 1 to 2 moles iron 0.2 1 mark -- mass iron 11.2 grams

Question 6 -- Concentration and Volume Calculations

Total: 12 marks

A technician at a laboratory in Birmingham prepares solutions of known concentration for use in titration experiments.

(a) [2]

(i) State the formula linking concentration (in mol/dm³), number of moles and volume (in dm³). [1]

(ii) Convert 250 cm³ to dm³. [1]

Model Answer -- 6(a)

concentration = moles / volume (in dm³) or c = n / V [1]

250 cm³ = 250 / 1000 = 0.250 dm³ [1]

⚠ If you missed marks here: There are 1000 cm³ in 1 dm³, so 250 cm³ = 0.250 dm³ – dividing by 100 (2.5 dm³) is the usual slip. The formula is concentration = moles ÷ volume in dm³; multiplying moles by volume, or using the volume in cm³, gives the wrong value.

1 mark -- concentration equals moles divided volume 1 mark -- 250 cm3 equals 0.250 dm3

(b) [3]

A technician dissolves 4.0 g of sodium hydroxide (NaOH) in water and makes the solution up to 500 cm³.

Calculate the concentration of the solution in mol/dm³.

Model Answer -- 6(b)

M_r of NaOH = 23 + 16 + 1 = 40; Moles = 4.0 / 40 = 0.1 mol [1]

Volume = 500 / 1000 = 0.5 dm³ [1]

Concentration = 0.1 / 0.5 = 0.2 mol/dm³ [1]

⚠ If you missed marks here: Change 500 cm³ into dm³ before dividing: 0.1 mol ÷ 0.5 dm³ = 0.2 mol/dm³, whereas dividing by 500 gives 0.0002. Also find the moles first (M_r of NaOH = 40) – 4.0 ÷ 0.5 = 8.0 is the concentration in g/dm³, not in mol/dm³.

1 mark -- moles NaOH 0.1 mol 1 mark -- volume 0.5 dm3 1 mark -- concentration 0.2 mol per dm3

The sodium hydroxide solution from part (b) is used in a titration with sulfuric acid:

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

It is found that 25.0 cm³ of the NaOH solution is exactly neutralised by 20.0 cm³ of the sulfuric acid.

Calculate the concentration of the sulfuric acid in mol/dm³.

Model Answer -- 6(c)

Moles NaOH = 0.2 × (25.0/1000) = 0.2 × 0.025 = 0.005 mol [1]

From equation: 2 mol NaOH reacts with 1 mol H₂SO₄, so moles H₂SO₄ = 0.005 / 2 = 0.0025 mol [1]

Volume H₂SO₄ = 20.0 / 1000 = 0.020 dm³ [1]

Concentration = 0.0025 / 0.020 = 0.125 mol/dm³ [1]

⚠ If you missed marks here: Two NaOH react with ONE H₂SO₄, so the acid moles are HALF the alkali moles: 0.005 ÷ 2 = 0.0025 mol – skipping this gives 0.25 mol/dm³, and doubling instead gives 0.50. Then divide by the ACID's volume, 20.0 cm³ = 0.020 dm³, not the 25.0 cm³ of NaOH.

1 mark -- moles NaOH 0.005 mol 1 mark -- mole ratio 2 to 1 moles H2SO4 0.0025 1 mark -- volume H2SO4 0.020 dm3 1 mark -- concentration 0.125 mol per dm3

(d) [3]

(i) State the formula to convert concentration from mol/dm³ to g/dm³. [1]

(ii) Calculate the concentration of the sulfuric acid solution from part (c) in g/dm³. [2]

Model Answer -- 6(d)

concentration (g/dm³) = concentration (mol/dm³) × M_r [1]

M_r of H₂SO₄ = (2×1) + 32 + (4×16) = 98 [1]

Concentration = 0.125 × 98 = 12.25 g/dm³ [1]

⚠ If you missed marks here: To go from mol/dm³ to g/dm³ you MULTIPLY by M_r (moles → grams): 0.125 × 98 = 12.25 g/dm³, whereas dividing gives 0.00128, far too small. Check the M_r: H₂SO₄ = 2 + 32 + 64 = 98; counting only one oxygen (50) is a common slip.

1 mark -- formula concentration mol dm3 times Mr 1 mark -- Mr H2SO4 equals 98 1 mark -- concentration 12.25 grams per dm3

Question 7 -- Gas Volumes at RTP

Total: 10 marks

Students at a school in Cambridge collect gases during experiments at room temperature and pressure (RTP).

(a) [2]

(i) State the molar volume of any gas at RTP. [1]

(ii) State the formula linking volume of gas, number of moles and molar volume. [1]

Model Answer -- 7(a)

The molar volume of any gas at RTP is 24 dm³/mol (or 24000 cm³/mol) [1]

volume = moles × molar volume or moles = volume / 24 [1]

⚠ If you missed marks here: At r.t.p. one mole of any gas occupies 24 dm³ (24 000 cm³) – 22.4 dm³ belongs to different conditions and is not used in this syllabus, and "24 cm³" is a thousand times too small. The formula is volume = moles × 24 (for dm³); dividing the moles by 24 is upside down.

1 mark -- molar volume 24 dm3 per mol 1 mark -- volume equals moles times molar volume

(b) [3]

Calcium carbonate reacts with dilute hydrochloric acid:

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

Calculate the volume of carbon dioxide gas produced at RTP when 10.0 g of calcium carbonate reacts with excess hydrochloric acid.

Model Answer -- 7(b)

Moles of CaCO₃ = 10.0 / 100 = 0.1 mol [1]

From equation: 1 mol CaCO₃ → 1 mol CO₂, so moles CO₂ = 0.1 mol [1]

Volume = 0.1 × 24 = 2.4 dm³ (or 2400 cm³) [1]

⚠ If you missed marks here: Follow the CaCO₃ : CO₂ ratio, which is 1 : 1, so 0.1 mol gives 0.1 mol of gas – using the 2 in front of HCl gives 4.8 dm³, which is wrong. Watch the units: 0.1 × 24 = 2.4 dm³ (2400 cm³), not 2.4 cm³.

1 mark -- moles CaCO3 0.1 mol 1 mark -- moles CO2 0.1 mol ratio 1 mark -- volume 2.4 dm3

Magnesium reacts with dilute sulfuric acid:

Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

A student collects 600 cm³ of hydrogen gas at RTP. Calculate the mass of magnesium that reacted.

Model Answer -- 7(c)

Moles of H₂ = 600 / 24000 = 0.025 mol (or 0.6 / 24 = 0.025) [1]

From equation: 1 mol Mg → 1 mol H₂, so moles Mg = 0.025 mol [1]

Mass of Mg = 0.025 × 24 = 0.60 g [1]

⚠ If you missed marks here: Convert the volume first: 600 ÷ 24 000 = 0.025 mol (or 0.600 dm³ ÷ 24) – dividing 600 by 24 gives 25 mol and an absurd 600 g of magnesium. Then 1 Mg : 1 H₂, so the mass is 0.025 × 24 = 0.60 g; the two 24s here mean different things (the A_r of Mg and the molar gas volume).

1 mark -- moles H2 0.025 mol 1 mark -- moles magnesium 0.025 ratio 1 mark -- mass magnesium 0.60 grams

(d) [2]

In the reaction: N₂(g) + 3H₂(g) → 2NH₃(g)

(i) State the ratio of volumes of nitrogen : hydrogen : ammonia in this reaction (at the same temperature and pressure). [1]

(ii) Calculate the volume of ammonia produced when 6.0 dm³ of nitrogen reacts with excess hydrogen. [1]

Model Answer -- 7(d)

Volume ratio N₂ : H₂ : NH₃ = 1 : 3 : 2 (same as mole ratio for gases at same T and P) [1]

Volume NH₃ = 6.0 × 2 = 12.0 dm³ [1]

⚠ If you missed marks here: For gases at the same temperature and pressure the numbers in the equation are VOLUME ratios, so 6.0 dm³ of N₂ gives 2 × 6.0 = 12.0 dm³ of NH₃ – there is no need for the 24 dm³ molar volume, and multiplying by it (144 dm³) is wrong. Halving instead (3.0 dm³) reads the ratio backwards.

1 mark -- volume ratio 1 to 3 to 2 1 mark -- volume ammonia 12 dm3

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