Chemistry 9701/34 — October/November 2015

Cambridge AS Level · Advanced Practical Skills 2 · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment you will determine the relative atomic mass, $A_r$ , of magnesium by a titration method.

FB 1 is $2.00\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ .
FB 3 is $0.120\text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$ .
magnesium ribbon
bromophenol blue indicator

(a)

Method

Reaction of magnesium with FB 1

Pipette $25.0\text{ cm}^3$ of FB 1 into the $250\text{ cm}^3$ beaker.
Weigh the strip of magnesium ribbon and record its mass.

$\text{mass of magnesium} = \text{................... g}$

Coil the strip of magnesium ribbon loosely and then add it to the FB 1 in the beaker.
Stir the mixture occasionally and wait until the reaction has finished.

Dilution of the excess acid

Transfer all the solution from the beaker into the volumetric flask.
Make the solution up to the mark using distilled water.
Shake the flask to mix the solution before using it for your titrations.
Label this solution of hydrochloric acid FB 2.

Titration

Fill the burette with FB 2.
Rinse the pipette out thoroughly. Then pipette $25.0\text{ cm}^3$ of FB 3 into a conical flask.
Add several drops of bromophenol blue indicator.
Perform a rough titration, by running the solution from the burette into the conical flask until the mixture just becomes yellow.
Record your burette readings in the space below.

$\text{The rough titre is ................... cm}^3.$

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make sure any recorded results show the precision of your practical work.
Record in a suitable form below all of your burette readings and the volume of FB 2 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

Record the mass of the magnesium ribbon, the rough titre, and all accurate burette readings.

Example readings:

mass of magnesium = 0.050 g
rough titre = 16.4 cm³

Accurate titrations:

initial reading / cm³	final reading / cm³	titre / cm³
0.00	16.35	16.35
0.00	16.35	16.35

All burette readings are recorded to the nearest 0.05 cm³. The two accurate titres are within 0.10 cm³ of each other.

Final answer

See working: example table with mass 0.050 g, rough titre 16.4 cm³, accurate titres 16.35 and 16.35 cm³

Detailed explanation

Background Concept

This is a back-titration experiment. Magnesium reacts with an excess of hydrochloric acid. The acid that is left unreacted is diluted to 250 cm³ and then titrated against sodium hydroxide. By finding how much acid remains, you can work out how much acid actually reacted with the magnesium, and hence the amount of magnesium. The quality of the final answer depends entirely on careful measurement and clear recording.

For titration results, burette readings must be recorded to the nearest 0.05 cm³ because a burette can be read to half a division. The titre is always final reading minus initial reading. Headings in a results table must include both the quantity and its unit, for example "titre / cm³".

Understanding the Question

Part (a) asks you to carry out the practical work and record the data. The mark scheme rewards four things: all necessary readings are present, the table has proper headings with units, readings are recorded to the correct precision, and at least two accurate titres agree within 0.10 cm³. Since this is a practical paper, the exact numbers are not known in advance; what matters is that the technique and presentation meet the required standard.

Approach

Weigh the magnesium ribbon before adding it to the acid. Carry out a rough titration first to find the approximate end-point, then repeat with care. For each accurate titration, record the initial and final burette readings, subtract to find the titre, and check that the concordant titres agree closely. Present the results in a clear table with units in the headings.

Step-by-Step Reasoning

Weigh the magnesium ribbon and record its mass. In this example, 0.050 g is used.
Perform a rough titration and record the rough titre, here 16.4 cm³.
Carry out accurate titrations. For each one:
- record the initial burette reading;
- add FB 2 until the indicator just turns yellow;
- record the final burette reading;
- calculate the titre as final reading minus initial reading.
Check that at least two accurate titres are within 0.10 cm³ of each other. Here both are 16.35 cm³, so they are concordant.
Present the data in a table with headings such as "initial reading / cm³", "final reading / cm³" and "titre / cm³".

The mark scheme also checks that no initial reading is 50.00 cm³ and that no reading exceeds 50.00 cm³.

Key Takeaways

Burette readings are recorded to the nearest 0.05 cm³.
Titre = final reading − initial reading.
A results table needs clear headings with units.
Concordant titres should agree within 0.10 cm³.

Common Mistakes

Writing "difference" or "total" instead of "titre" or "volume of FB 2 added".
Omitting units from table headings.
Recording burette readings to 1 decimal place; they must be to 2 decimal places, ending in 0 or 5.
Using 50.00 as an initial burette reading.
Performing only one accurate titration.

Things to Be Careful About

The mark scheme rejects readings greater than 50.00 cm³.
If a third titration is outside 0.10 cm³ of the first two, concordance may be lost unless another titre is also within 0.10 cm³ of one of them.
Keep the same number of decimal places throughout the table.

Techniques used

record the mass of magnesiumrecord initial and final burette readingscalculate titres by subtractiontabulate titration results with headings and unitscheck concordance of accurate titres

(b)

From your accurate titration results, obtain a suitable value for the volume of FB 2 to be used in your calculations.
Show clearly how you have obtained this value.

$25.0\text{ cm}^3$ of FB 3 required ................... $\text{cm}^3$ of FB 2.

DifficultyEasy

Worked solution

Working

The two accurate titres are 16.35 cm³ and 16.35 cm³.

\text{mean titre} = \frac{16.35 + 16.35}{2} = 16.35\text{ cm}^3

Answer

25.0 cm³ of FB 3 required 16.35 cm³ of FB 2.

Final answer

16.35 cm³

Detailed explanation

Background Concept

When several titres are concordant, the most reliable value to use in calculations is their mean. The mean is found by adding the selected titres and dividing by the number of titres used. It should be quoted to two decimal places, rounded to the nearest 0.01 cm³.

Understanding the Question

Part (b) asks you to choose a suitable value for the volume of FB 2 used in the calculations. You must show which titres you selected and how you averaged them. The mark scheme requires the selected titres to be within a total spread of no more than 0.20 cm³.

Approach

Look at the accurate titres from part (a), select two or more that agree closely, and calculate their mean. Show the addition and division clearly.

Step-by-Step Reasoning

The accurate titres recorded are 16.35 cm³ and 16.35 cm³.
They are identical, so their spread is 0.00 cm³, well within 0.20 cm³.
Mean = (16.35 + 16.35) / 2 = 16.35 cm³.
The mean is already to two decimal places, so no further rounding is needed.

The selected titres should be ticked or otherwise identified so the examiner can see which ones were used.

Key Takeaways

Use only concordant titres when calculating a mean.
Show the averaging working.
Quote the mean to two decimal places unless a special case applies.

Common Mistakes

Averaging all titres, including an obvious rough or anomalous value.
Quoting the mean to too many decimal places.
Not showing which titres were selected.

Things to Be Careful About

If all accurate readings were recorded to 1 decimal place, the mean may be quoted to 1 decimal place only if it is exactly correct.
A mean ending in 0.025 or 0.075 may be quoted to 3 decimal places.

Techniques used

select concordant titrescalculate the mean titre

(c)

Calculations

Show your working and appropriate significant figures in the final answer to each step of your calculations.

(i)

Calculate the number of moles of sodium hydroxide present in $25.0\text{ cm}^3$ of solution FB 3.

$\text{moles of NaOH} = \text{................... mol}$

DifficultyEasy

Worked solution

Working

\text{moles NaOH} = 0.120 \times \frac{25.0}{1000} = 0.00300\text{ mol}

Answer

moles of NaOH = 0.00300 mol

Final answer

0.00300 mol

Detailed explanation

Background Concept

The amount of a solute in moles is found from its concentration and the volume of solution:

\text{moles} = \text{concentration} \times \text{volume in dm}^3

Since volumes in titration are usually given in cm³, they must be converted to dm³ by dividing by 1000.

Understanding the Question

FB 3 is 0.120 mol dm⁻³ NaOH. You pipette 25.0 cm³ of it, so you need the number of moles of NaOH in that portion. This value is used later to find the amount of HCl in the titre.

Approach

Substitute the concentration and converted volume into the moles formula.

Step-by-Step Reasoning

Volume = 25.0 cm³ = 25.0 / 1000 = 0.0250 dm³.
Moles = 0.120 × 0.0250 = 0.00300 mol.

The answer is kept to three significant figures because the concentration has three significant figures.

Key Takeaways

Always convert cm³ to dm³ before using concentration in mol dm⁻³.
Moles = concentration × volume.

Common Mistakes

Forgetting to divide the volume by 1000.
Writing 0.003 mol instead of 0.00300 mol when three significant figures are intended.

Things to Be Careful About

The units of concentration are mol dm⁻³, so volume must be in dm³.

Techniques used

calculate moles from concentration and volume

(ii)

Give the equation for the reaction of hydrochloric acid, $\text{HCl}$ , with sodium hydroxide, $\text{NaOH}$ . State symbols are not required.

Deduce the number of moles of hydrochloric acid in the volume of FB 2 you calculated in (b).

$\text{moles of }\text{HCl} = \text{................... mol}$

DifficultyEasy

Worked solution

Answer

\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}

Since the reaction is 1:1, the moles of HCl in the titre of FB 2 are equal to the moles of NaOH in 25.0 cm³ of FB 3.

moles of HCl = 0.00300 mol

Final answer

0.00300 mol

Detailed explanation

Background Concept

Hydrochloric acid and sodium hydroxide react in a 1:1 mole ratio:

\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

One mole of acid neutralises exactly one mole of alkali. Therefore, at the end-point, the moles of HCl that have reacted equal the moles of NaOH originally present.

Understanding the Question

You must write the equation for the neutralisation and then use it to find the moles of HCl in the volume of FB 2 that reacted with the 25.0 cm³ of FB 3. The volume of FB 2 is the mean titre from part (b), 16.35 cm³.

Approach

Write the balanced equation, identify the 1:1 ratio, and copy the moles of NaOH from part (c)(i) as the moles of HCl.

Step-by-Step Reasoning

The balanced equation is NaOH + HCl → NaCl + H₂O.
The coefficients show 1 mol NaOH reacts with 1 mol HCl.
Moles of NaOH in the conical flask = 0.00300 mol.
Therefore moles of HCl in the titre = 0.00300 mol.

Key Takeaways

Acid–base neutralisation of a monoprotic acid with a monobasic base is 1:1.
The moles of one reactant can be transferred directly when the stoichiometric ratio is 1:1.

Common Mistakes

Writing an unbalanced equation.
Using a 2:1 ratio because HCl contains one H and NaOH contains one OH.
Including state symbols when the question says they are not required; they are not penalised but are unnecessary.

Things to Be Careful About

The answer to part (ii) must match the answer to part (i).

Techniques used

write the neutralisation equationdeduce moles from 1:1 stoichiometry

(iii)

Calculate the number of moles of hydrochloric acid in $250\text{ cm}^3$ of FB 2.

$\text{moles of }\text{HCl}\text{ in } 250\text{ cm}^3\text{ of }\text{FB 2} = \text{................... mol}$

DifficultyMedium-Easy

Worked solution

Working

The titre volume is 16.35 cm³, so:

\text{moles HCl in } 250\text{ cm}^3 = 0.00300 \times \frac{250}{16.35} = 0.0459\text{ mol}

Answer

moles of HCl in 250 cm³ of FB 2 = 0.0459 mol

Final answer

0.0459 mol

Detailed explanation

Background Concept

FB 2 is a homogeneous solution, so the concentration is the same in every portion. If a 16.35 cm³ portion contains 0.00300 mol, then a 250 cm³ portion contains proportionally more moles. The scaling factor is the ratio of the volumes.

Understanding the Question

Part (c)(ii) gave the moles of HCl in the titre volume, 16.35 cm³. The whole of FB 2 was made up to 250 cm³, so you need the total moles of unreacted acid in the whole flask.

Approach

Multiply the moles in the titre by the ratio of the total volume to the titre volume.

Step-by-Step Reasoning

Moles in 16.35 cm³ = 0.00300 mol.
Total volume = 250 cm³.
Moles in 250 cm³ = 0.00300 × (250 / 16.35) = 0.0459 mol.

The result is given to three significant figures, consistent with the data.

Key Takeaways

A volume ratio can scale a quantity of solute from one portion to the whole solution.
The ratio must have the same units in the numerator and denominator so they cancel.

Common Mistakes

Inverting the volume ratio.
Forgetting that FB 2 is the diluted excess acid, not the original FB 1.
Rounding too early and losing precision.

Things to Be Careful About

Use the volume from part (b), not the rough titre.
Keep enough significant figures during the calculation before rounding the final answer.

Techniques used

scale moles by volume ratio

(iv)

Calculate the number of moles of hydrochloric acid in $25.0\text{ cm}^3$ of FB 1.

$\text{moles of }\text{HCl}\text{ in } 25.0\text{ cm}^3\text{ of }\text{FB 1} = \text{................... mol}$

DifficultyEasy

Worked solution

Working

\text{moles HCl in } 25.0\text{ cm}^3\text{ of FB 1} = 2.00 \times \frac{25.0}{1000} = 0.0500\text{ mol}

Answer

moles of HCl in 25.0 cm³ of FB 1 = 0.0500 mol

Final answer

0.0500 mol

Detailed explanation

Background Concept

FB 1 is the original 2.00 mol dm⁻³ hydrochloric acid. The 25.0 cm³ portion placed in the beaker contains a fixed number of moles of HCl, calculated from concentration and volume.

Understanding the Question

This step finds the total acid available before any reaction with magnesium. It will be compared with the acid remaining after the reaction to find how much was used up.

Approach

Use moles = concentration × volume, converting 25.0 cm³ to dm³.

Step-by-Step Reasoning

Volume = 25.0 cm³ = 0.0250 dm³.
Moles = 2.00 × 0.0250 = 0.0500 mol.

Key Takeaways

The original acid amount is found in the same way as any solution amount.
This value is the "starting" amount of HCl before reaction.

Common Mistakes

Using the diluted concentration of FB 2 instead of FB 1.
Forgetting to convert cm³ to dm³.

Things to Be Careful About

FB 1 is 2.00 mol dm⁻³; FB 2 is the diluted solution after reaction.

Techniques used

calculate moles from concentration and volume

(v)

In (a), you reacted $25.0\text{ cm}^3$ of FB 1 with your weighed piece of magnesium. After the reaction, the unreacted hydrochloric acid was used to prepare $250\text{ cm}^3$ of FB 2.

Use your answers to (iii) and (iv) to calculate the number of moles of hydrochloric acid that reacted with the magnesium ribbon.

$\text{moles of }\text{HCl}\text{ reacting with Mg} = \text{................... mol}$

DifficultyEasy

Worked solution

Working

\text{moles HCl reacting with Mg} = 0.0500 - 0.0459 = 0.0041\text{ mol}

Answer

moles of HCl reacting with Mg = 0.0041 mol

Final answer

0.0041 mol

Detailed explanation

Background Concept

The acid placed with the magnesium is partly consumed by the reaction. The acid left over is what was diluted to make FB 2. Therefore:

acid used by magnesium = acid originally present − acid remaining

Understanding the Question

Part (c)(iv) gave the acid originally present in 25.0 cm³ of FB 1. Part (c)(iii) gave the acid remaining after reaction, in the 250 cm³ of FB 2. Subtracting gives the acid that reacted with magnesium.

Approach

Subtract the moles from part (iii) from the moles from part (iv).

Step-by-Step Reasoning

Original HCl = 0.0500 mol.
Remaining HCl = 0.0459 mol.
HCl reacted = 0.0500 − 0.0459 = 0.0041 mol.

The answer has two significant figures because the subtraction leaves a small difference; the limiting precision is around two significant figures.

Key Takeaways

Back-titration logic: initial amount − final amount = amount reacted.
Subtraction can reduce the number of significant figures reliably quoted.

Common Mistakes

Adding the two amounts instead of subtracting.
Using the titre moles directly as the reacted acid.
Quoting too many significant figures after a subtraction.

Things to Be Careful About

The result must be positive; if it is negative, the volumes or calculations have been inverted.

Techniques used

subtract remaining acid from initial acid

(vi)

Complete the equation below, for the reaction of magnesium with hydrochloric acid.
State symbols are required.

$\text{Mg} \quad + \quad \text{HCl} \rightarrow \text{MgCl}_2 \quad + \quad \text{............................}$

Use your answer to (v) to calculate the number of moles of magnesium used.

$\text{moles of Mg} = \text{................... mol}$

DifficultyMedium-Easy

Worked solution

Answer

\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}

From the equation, 2 mol HCl react with 1 mol Mg:

\text{moles Mg} = \frac{0.0041}{2} = 0.00205\text{ mol}

moles of Mg = 0.00205 mol

Final answer

0.00205 mol

Detailed explanation

Background Concept

Magnesium is a Group 2 metal. It reacts with hydrochloric acid to form magnesium chloride and hydrogen gas:

\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2

The balanced equation shows that two moles of HCl are needed for every one mole of Mg. Therefore the moles of Mg are half the moles of HCl that reacted.

Understanding the Question

You must complete the equation with the missing product and state symbols, then use the moles of HCl from part (v) to find the moles of magnesium.

Approach

Identify hydrogen as the other product, balance the equation, and divide the moles of HCl by 2.

Step-by-Step Reasoning

Magnesium reacts with acid to give a salt and hydrogen: MgCl₂ and H₂.
Balancing: one Mg gives one MgCl₂; two HCl provide two Cl and two H, so the equation is Mg + 2HCl → MgCl₂ + H₂.
State symbols: Mg(s), HCl(aq), MgCl₂(aq), H₂(g).
Moles of HCl reacted = 0.0041 mol.
Moles of Mg = 0.0041 / 2 = 0.00205 mol.

Key Takeaways

Group 2 metals react with acids to produce hydrogen gas.
The stoichiometric ratio from the balanced equation is essential for converting moles of one substance to moles of another.

Common Mistakes

Writing MgCl instead of MgCl₂.
Forgetting the state symbols when they are required.
Dividing by the wrong ratio, e.g. using 1:1 instead of 2:1.

Things to Be Careful About

The question explicitly says state symbols are required here, unlike part (c)(ii).
Use the unrounded or properly rounded value from part (v) to avoid excessive rounding error.

Techniques used

balance the equation for magnesium with hydrochloric aciduse stoichiometric ratio to find moles of magnesium

(vii)

Use your answer to (vi) to calculate the relative atomic mass, $A_r$ , of magnesium.

$A_r\text{ of Mg} = \text{...................}$

DifficultyMedium-Easy

Worked solution

Working

A_r = \frac{\text{mass of Mg}}{\text{moles of Mg}} = \frac{0.050}{0.00205} = 24.4

Answer

A_r of Mg = 24.4

Final answer

24.4

Detailed explanation

Background Concept

The relative atomic mass is the mass of one mole of atoms. If you know the mass of a sample and the number of moles in it, then:

A_r = \frac{\text{mass in g}}{\text{amount in mol}}

The units are grams per mole, numerically equal to the relative atomic mass.

Understanding the Question

You have the mass of magnesium recorded in part (a) and the moles of magnesium from part (c)(vi). Dividing mass by moles gives the experimental A_r.

Approach

Substitute the recorded mass and the calculated moles into the formula.

Step-by-Step Reasoning

Mass of Mg = 0.050 g.
Moles of Mg = 0.00205 mol.
A_r = 0.050 / 0.00205 = 24.4.

The accepted value for magnesium is about 24.3, so this experimental result is close.

Key Takeaways

A_r can be found experimentally from mass and moles.
The calculation is a simple division, but the accuracy depends on every previous step.

Common Mistakes

Using the mass of the whole experiment incorrectly.
Using moles of HCl instead of moles of Mg.
Quoting the answer without considering significant figures.

Things to Be Careful About

The mass must be in grams.
If the mass was recorded to two significant figures, the final A_r should not claim more precision than the data allow.

Techniques used

calculate relative atomic mass from mass and moles

(d)

(i)

State one observation that proves that the hydrochloric acid in FB 1 was in excess for the reaction with the magnesium ribbon.

DifficultyEasy

Worked solution

Answer

All of the solid magnesium dissolved / disappeared / reacted.

Alternatively: when FB 2 was added to the alkali during the titration, the indicator turned from blue to yellow, showing that acid was present in excess.

Final answer

All solid magnesium dissolved / indicator turned from blue to yellow when FB 2 was added

Detailed explanation

Background Concept

For a back-titration to work, the acid must be in excess so that some acid remains after the magnesium has fully reacted. The remaining acid is then titrated. If the acid were not in excess, all the acid would be used up and there would be nothing left to titrate.

Understanding the Question

The question asks for one observation from the procedure that proves the acid was in excess. You need to link a visible observation to the chemical requirement that acid remained after the reaction.

Approach

Think about what you would see if the magnesium reacted completely and acid was still present. The solid magnesium disappearing shows the metal was the limiting reactant and was used up. Also, the titration end-point with bromophenol blue going from blue to yellow shows that acidic solution was being added to alkali, so acid was available.

Step-by-Step Reasoning

If the magnesium ribbon had not all reacted, solid metal would remain in the beaker.
Seeing that all the solid magnesium dissolved shows the magnesium was completely consumed.
Since the reaction stopped while acid was still present, the acid must have been in excess.
Alternatively, during the titration, the indicator changing from blue to yellow when FB 2 is added shows the acidic FB 2 is neutralising the alkali; this also confirms acid was present in excess after the magnesium reaction.

Key Takeaways

A back-titration requires the reagent being analysed to be in excess.
Observations such as complete dissolution of a solid can prove which reactant is limiting.

Common Mistakes

Saying "bubbles were produced" — this shows reaction occurred but not that acid was in excess.
Saying "the solution turned yellow" without linking it to the titration with FB 2.

Things to Be Careful About

The observation must be one that could actually be seen in the procedure described.

Techniques used

identify an observation indicating complete reaction and excess acid

(ii)

A student carried out exactly the same experiment but used $1.00\text{ g}$ of magnesium ribbon.

State and explain why the student's experiment could not be used to determine the value for the $A_r$ of magnesium.
Include a calculation in your answer.

DifficultyMedium

Worked solution

Answer

The experiment would fail because the magnesium would be in excess and the acid would be the limiting reagent; all the hydrochloric acid would be used up, leaving no excess acid to titrate.

Calculation:

\text{moles Mg} = \frac{1.00}{24.3} = 0.041\text{ mol}

\text{moles HCl needed} = 2 \times 0.041 = 0.082\text{ mol}

But only 0.0500 mol of HCl is available, so the acid is used up completely and the back-titration cannot be carried out.

Final answer

Mg in excess / acid limiting; calculation shows 0.082 mol HCl needed but only 0.0500 mol available

Detailed explanation

Background Concept

In a back-titration, the substance being analysed must be the limiting reagent, so that an excess of the titrated reagent remains. Here, the acid must be in excess relative to the magnesium. If too much magnesium is used, the acid is completely consumed and there is no unreacted acid to dilute and titrate.

Understanding the Question

A student uses 1.00 g of magnesium instead of the smaller mass used in the experiment. You must state why this prevents determination of A_r and support your answer with a calculation comparing the moles of acid available with the moles needed.

Approach

Calculate the moles of Mg in 1.00 g, then use the 2:1 HCl:Mg ratio to find the moles of HCl required. Compare this with the 0.0500 mol of HCl actually present in 25.0 cm³ of 2.00 mol dm⁻³ acid.

Step-by-Step Reasoning

Moles of Mg = mass / A_r = 1.00 / 24.3 = 0.041 mol.
From the equation Mg + 2HCl → MgCl₂ + H₂, each mole of Mg needs 2 moles of HCl.
Moles of HCl needed = 2 × 0.041 = 0.082 mol.
Moles of HCl available = 2.00 × 25.0/1000 = 0.0500 mol.
Since 0.082 mol > 0.0500 mol, the acid is insufficient; the magnesium is in excess.
All the acid reacts with some of the magnesium, leaving no excess acid to dilute and titrate.
Without knowing how much acid remained, you cannot find how much acid reacted, so you cannot determine the moles of magnesium and hence A_r.

Key Takeaways

The limiting reagent concept is central to back-titration design.
A stoichiometric calculation can prove whether a proposed mass is suitable.
If the acid is limiting, the back-titration method collapses.

Common Mistakes

Forgetting to multiply moles of Mg by 2 when finding HCl needed.
Comparing masses instead of moles.
Stating that the magnesium would be in excess without showing the calculation.

Things to Be Careful About

The calculation must be included for full marks.
Use the A_r of magnesium from part (c)(vii) if preferred, or the accepted value 24.3.

Techniques used

compare moles of reactants to identify the limiting reagentcalculate moles from mass and relative atomic mass

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation9M
Q3Qualitative Analysis · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation14M

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