Edexcel A-Level Chemistry AS Paper 2, June 2018: Question 2

12 marks · Medium difficulty · Calculations

Calculate gas volumes using the ideal gas equation and molar volume at r.t.p., and determine reacting quantities and excess reagents from a titration and gas evolution experiment.

Practise this question

Question

Question text

2 This question is about the molar volume of gases.

(a) (i) Calculate the volume of one mole of an ideal gas, A, at 60°C and 500kPa pressure.

Give your answer to two significant figures and include units.

[The ideal gas equation is pV = nRT. Gas constant (R) = 8.31 J K−1 mol−1]

(3)

(ii) At room temperature and pressure (r.t.p) another gas B, with formula XH3,

has a density of 1.42 g dm−3.

Calculate the molar mass of the gas XH3 and deduce the identity of the

element X.

[The molar volume of gas B = 24 000 cm3 mol−1 at r.t.p.]

(2)

(b) The apparatus shown was used to measure the volume of gas evolved when a

weighed mass of sodium carbonate reacted with dilute hydrochloric acid.

delivery tube

rubber bung

syringe

test tube

The following procedure was used.

Step 1 Solid sodium carbonate was placed in a container and weighed accurately.

Step 2 The delivery tube and rubber bung were removed and the

sodium carbonate was transferred to the test tube.

Step 3 The container was then reweighed.*P51460RA0324*

Step 4 The syringe plunger was pushed in, to zero the syringe.

Step 5 10.0 cm3 of 0.400 mol dm−3 hydrochloric acid was then added to the

sodium carbonate and the rubber bung and delivery tube rapidly replaced.

Step 6 The mixture was shaken and, when the reaction had finished,

the reading of the syringe was noted.

Results

Mass of container and sodium carbonate before transfer = 20.135g

Mass of container after transfer of the sodium carbonate = 19.893g

Mass of sodium carbonate used = 0.242g

The equation for the reaction is

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

(i) Calculate the moles of hydrochloric acid and the moles of sodium carbonate

used in this experiment.

Use your answers to decide which reactant is in excess.

Calculate the maximum volume of carbon dioxide which could be produced.

Molar mass of Na CO = 106.0 g mol−1

3 −1 (5)

Molar volume of gas = 24000cm mol at r.t.p.

*P51460RA0424*

(ii) The actual volume of carbon dioxide collected was less than calculated.

Give two reasons for this.

(2)

(Total for Question 2 = 12 marks)

Mark scheme

Show the mark scheme

Question

Acceptable Answer Additional Guidance Mark

Number

2 (a)(i) Examples of calculation

60 oC = 333 K

converts temperature to Kelvin and pressure to

-2 500 kPa = 5 x 105 / 500 000 Pa

Nm (Pa) (1)

rearranging ideal gas equation and substituting V = nRT

their values (1) P

V = 1 x 8.31 x 333/500 000

evaluates answer to 2 SF and includes units (1) -3

= 5.53446 x 10

= 0.0055 m3/5.5 x 10-3 m3 / 5.5 dm3 / 5500 cm3

allow TE

answers to 2 SF only

correct answer with no working scores 3 marks

correct answer with incorrect working scores 2

marks max. (3)

Question

Acceptable Answer Additional Guidance Mark

Number

2(a)(ii) Example of calculation:

calculates M to 2 or more SF (1) molar mass = mass in 24000 cm3

= 1.42 x 24000/1000 = 34 (.08) (g mol-1)

ignore SF except 1 SF

identifies element X (1) (X + (3 x 1)) = 34

X = 31 so P / phosphorus

just ‘phosphorus’ with no working scores M2

only (2)

Question Acceptable Answer Additional Guidance Mark

Number

2(b)(i) Example of calculation

calculates moles of acid (1) moles of acid =10.0 x 0.400/1000

=4(.0) x 10-3 /0.004 (mol)

calculates moles of sodium carbonate (1) moles of sodium carbonate =0.242/106.0

= 2.283 x 10-3/0.002283 (mol)

recognises that (sodium) carbonate is in excess

(1)

evidence for excess sodium carbonate in terms of recognition of HCl:Na2CO3 = 2:1 gets M4

moles (1) 4.0 x10-3 mol acid requires

2.0 x 10-3 mol sodium carbonate

2.283 x 10-3 mol of sodium carbonate requires

4.566 x 10-3 mol of acid

correct volume of gas calculated with units (1) moles CO = 2.0 x 10-3 (mol)

volume of gas = 2.0 x 10-3 x 24 000

= 48 cm3 /0.048 dm3

TE on incorrect moles CO2

correct answer with no working scores 1 mark

if the moles of sodium carbonate are not

calculated, only M1, M4 and M5 can be

awarded.

ignore SF except 1 for M5 (5)

Question Acceptable Answer Additional Guidance Mark

Number

2(b)(ii) An answer that makes reference to the following

reasons:

ignore references to change in volume when

some gas escaped before the bung/delivery tube the bung is pushed into the test tube

was replaced (1)

allow ‘temperature less than

the gas / carbon dioxide is (slightly) soluble in 25oC/298 K/room temperature’ as alternative

water/ acid / solution (1) to either answer

do not award an incomplete reaction

do not award leaky apparatus/sticking syringe (2)

(Total for Question 2 = 12 marks)

How to answer it

The Molar Volume of Gases & Stoichiometry

What this question tests

This multi-part exam question assesses core physical chemistry competencies: rearranging and applying the ideal gas equation ( pV = nRT ), understanding molar volume at r.t.p., performing multi-step titration/solid stoichiometry calculations (limiting reagents), and critically evaluating experimental errors in volumetric gas analysis.

Question 2 (a) (i)

Ideal Gas Equation Calculation

💡 Key Knowledge

Units must be converted to SI units before substitution: Temperature to Kelvin ( °C + 273 ), Pressure to Pascals ( Pa or N m⁻² ).
Rearranging pV = nRT for volume V gives V = nRT / p .
For 1 mole of gas, n = 1 .

📐 Step-by-Step Calculation

Convert units:
T = 60 + 273 = 333 K
p = 500 kPa = 500 × 10³ Pa (or 500 000 Pa )
Substitute into rearranged equation:
V = (1 × 8.31 × 333) / 500 000
Evaluate:
V = 0.005534... m³ (or 5.5 dm³ / 5500 cm³ )

✅ Correct Answer & Mark Scheme

Answer: 0.0055 m³ (or 5.5 dm³ / 5500 cm³ )

Mark Breakdown (3 marks):
1 mark for converting temperature to Kelvin and pressure to Pa.
1 mark for correct rearrangement and substitution.
1 mark for final evaluation to 2 significant figures with correct units.

❌ Common Errors & Traps

Forgetting to convert kPa to Pa (missing the ×10³ factor).
Failing to give the final answer to the requested 2 significant figures.
Omitting or stating incorrect volumetric units.

Question 2 (a) (ii)

Molar Mass and Elemental Deduction

💡 Key Knowledge

At room temperature and pressure (r.t.p.), 1 mole of any gas occupies 24 000 cm³ (or 24 dm³ ). Density is mass per unit volume ( g dm⁻³ ).

📐 Step-by-Step Calculation

Calculate molar mass ( Mᵣ ):
Mᵣ = density × molar volume
Mᵣ = 1.42 × 24 = 34.08 g mol⁻¹ ( 34 to 2 SF)
Deduce element X:
Formula is XH₃ , so Mᵣ(X) + (3 × 1.0) = 34
Ar(X) = 34 - 3 = 31
Looking up Ar = 31 on the Periodic Table identifies X as Phosphorus (P).

✅ Correct Answer & Mark Scheme

Answer: Molar mass = 34 g mol⁻¹ , Element X = Phosphorus / P

Mark Breakdown (2 marks):
1 mark for calculating Mᵣ to 2 or more SF.
1 mark for identifying element X (name or symbol). Note: Just stating 'phosphorus' without working scores M2 only.

Question 2 (b) (i)

Stoichiometry, Limiting Reagents & Gas Volume

💡 Key Knowledge

To find the limiting reagent, compare the calculated moles of reactants against the molar stoichiometric ratio from the balanced equation: Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + CO₂(g) + H₂O(l)

📐 Step-by-Step Calculation

Moles of acid ( HCl ):
n = (10.0 / 1000) × 0.400 = 4.0 × 10⁻³ mol
Moles of sodium carbonate ( Na₂CO₃ ):
Mass used = 20.135 - 19.893 = 0.242 g
n = 0.242 / 106.0 = 2.283 × 10⁻³ mol
Determine limiting reagent:
Ratio is 1 mol Na₂CO₃ : 2 mol HCl .
4.0 × 10⁻³ mol of acid requires 2.0 × 10⁻³ mol of Na₂CO₃ . Since we have 2.283 × 10⁻³ mol of Na₂CO₃ available, Na₂CO₃ is in excess, making HCl the limiting reagent.
Calculate maximum CO₂ volume:
From 2 mol HCl → 1 mol CO₂ :
n(CO₂) = 4.0 × 10⁻³ / 2 = 2.0 × 10⁻³ mol
Volume = 2.0 × 10⁻³ × 24 000 = 48 cm³ (or 0.048 dm³ )

✅ Correct Answer & Mark Scheme

Answer: 48 cm³ (or 0.048 dm³ )

Mark Breakdown (5 marks):
1 mark: Moles of acid calculated correctly.
1 mark: Moles of sodium carbonate calculated correctly.
1 mark: Recognises sodium carbonate is in excess.
1 mark: Evidence for excess in terms of mole comparison.
1 mark: Correct final volume of gas with units.

🧠 Exam Technique & Common Traps

Students frequently trip up by basing their gas volume calculation on the excess reagent instead of the limiting reagent ( HCl ). Always use the limiting reagent to determine theoretical yields.

Question 2 (b) (ii)

Evaluating Experimental Discrepancies

✅ Acceptable Answers (Any two)

Some gas escaped before the bung/delivery tube was replaced (Step 5 delay).
The carbon dioxide gas is slightly soluble in water/acid/solution.
Temperature of the room was lower than standard r.t.p. ( 25 °C ).

Mark Breakdown (2 marks): 1 mark per valid reason (max 2).
Note: Ignore vague answers like "leaky apparatus" or "incomplete reaction" unless specifically tied to gas escape or solubility.

💡 Examiner Insight

Top-level responses immediately spot practical flaws related to apparatus setup timing (gas loss during addition of acid before sealing) and physical properties of the gas evolved ( CO₂ solubility in aqueous media).

Topics

Physical Chemistry · Core Practicals · Topic 5: Formulae, Equations and Amounts of Substance · Core Practical 1: Measuring the molar volume of a gas

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.