Limiting Reactants [HT] and Concentration of Solutions

AQA 8462 — 4.3.2.4 & 4.3.2.5

⬢ HIGHER TIER PARTIAL — LIMITING REACTANTS HT ONLY ⬢

Limiting Reactants & Concentration of Solutions

Two quantitative concepts that decide how much product you can make and how much solute you've packed into your solution. Both turn on careful mole calculations and one critical unit conversion.

What You'll Learn

Part 1 — HT Only

Limiting reactants: identify which reactant runs out first, explain how it determines the maximum mass or moles of product, and recognise why an excess reactant is sometimes used deliberately (e.g. RP1).

Part 2 — Both Tiers

Concentration of solutions in g/dm³: apply the formula c = m ÷ V, perform the critical cm³ → dm³ conversion, and (HT only) explain how concentration is related to mass of solute and volume of solution.

Why It Matters

These are calculation topics. Marks come from labelled steps, correct unit conversions, and naming the limiting reactant rather than describing it vaguely.

The Limiting Reactant

When two reactants are mixed, they're rarely present in the exact ratio the equation demands. One runs out first — and once it's gone, the reaction stops producing more product, no matter how much of the other reactant is left.

Limiting Reactant

The reactant that is used up completely in a reaction — it determines the maximum amount of product that can form.

In Excess

A reactant present in more than the amount needed to react with the limiting reactant. Some of it is left over at the end of the reaction.

How to Identify the Limiting Reactant

Don't guess by mass — masses don't tell you about the ratio in the equation. Always work in moles.

Step 1Calculate the moles of each reactant using moles = mass ÷ M_r.

Step 2Use the mole ratio from the balanced equation to check which reactant runs out first.

Step 3That reactant is the limiting reactant. Calculate moles or mass of product from it.

The Test

For a balanced equation A + 2B → C: if you have 1 mol A and 1 mol B, then A would need 2 mol B to fully react. You only have 1 mol B → B runs out first → B is limiting. Product formed depends on B (1 mol B → 0.5 mol C), not on A.

Worked Example 1

1

Identify the Limiting Reactant HT

In the reaction Mg + 2HCl → MgCl₂ + H₂, a student mixes 0.10 mol of Mg with 0.15 mol of HCl. Identify the limiting reactant. Calculate the moles of MgCl₂ produced. [3 marks]

Step 1 — Mole ratio: Mg : HCl = 1 : 2 (from the equation)

Step 2 — Test Mg: 0.10 mol Mg would need 0.10 × 2 = 0.20 mol HCl

We only have 0.15 mol HCl → not enough

Step 3 — Test HCl: 0.15 mol HCl would react with 0.15 ÷ 2 = 0.075 mol Mg

We have 0.10 mol Mg → Mg is in excess

∴ HCl is the limiting reactant. Moles of MgCl₂ = 0.15 ÷ 2 = 0.075 mol

Examiner Alert — Name the Substance

Real exam context (P2 Jun22 Q08.4, RPA5): a rate graph for HCl + sodium thiosulfate (thiosulfate in excess) plateaus at ~80 seconds. The 1-mark question: "Why did the reaction stop?"

Examiner Alert — P2 Jun22 Q08.4 [1 mark]

The mark scheme awards the mark for "(hydrochloric) acid is used up" or "(hydrochloric) acid is the limiting reactant". It explicitly says: "ignore reactants used up" — saying "the reactants were used up" without naming the specific substance scored zero.

Loses the mark

"The reactants were used up."

"They both ran out."

"The limiting reactant was used up." (without naming it)

Earns the mark

"The hydrochloric acid was used up."

"Hydrochloric acid was the limiting reactant."

"All of the HCl had reacted."

Exam Tip

The concept is about one specific substance running out. Name it. Use a chemical name, not a pronoun ("it" loses marks too).

Excess in RP1: Why the Insoluble Reactant?

In RP1 (preparing a soluble salt), the insoluble reactant is added in excess — e.g. excess copper oxide added to dilute sulfuric acid, or excess zinc oxide added to dilute hydrochloric acid. The exam question isn't whether to use excess; it's why excess of the insoluble reactant rather than excess of the acid.

The Discriminator — Removability

Both options would use up all of the other reactant. The difference is what happens to the leftover excess at the end:

Excess insoluble base (CuO, ZnO) → stays as a solid → can be filtered off (residue) → leaves a pure salt solution (filtrate).

Excess acid → dissolved in solution → cannot be separated by filtration → would contaminate the salt.

Examiner Alert — P1 Jun22 Q04.2 [1 mark]

The actual question: "Why is excess zinc oxide used rather than excess hydrochloric acid?" The mark scheme awards the mark for "(excess) zinc oxide can be filtered off" (or "easier to remove / separate"). It explicitly ignores answers along the lines of "to use up all the (hydrochloric) acid" — that reasoning doesn't distinguish between the two options. Examiner report: "Few students gained this mark... answers in terms of ease of removal of the excess reactant were required."

Exam Tip

For "why excess of the insoluble reactant?" → focus on filtration / removal. The fact that "all the acid reacts" is true of either option, so it doesn't earn the mark. Also note: 4.3.2.4 (HT) does state excess is used "to ensure all of the other reactant is used" — that's the spec concept; Q04.2 marks the procedural reason.

Concentration in g/dm³

Many chemical reactions take place in solutions. To compare them quantitatively we need a way to say how much solute is dissolved per unit volume.

Concentration

The amount of solute dissolved in a given volume of solution. The standard GCSE unit is grams per decimetre cubed (g/dm³).

Volume Conventions

1 dm³ = 1000 cm³ = 1 litre. The decimetre cubed is the same volume as a litre — exam questions usually quote volumes in cm³, so you'll almost always need to convert.

The Formula

concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)

Critical Step — Always Convert Volume First

Volume in the formula must be in dm³, not cm³. To convert:volume in dm³ = volume in cm³ ÷ 1000Examples: 50 cm³ = 0.05 dm³; 250 cm³ = 0.25 dm³; 1000 cm³ = 1.0 dm³.

Cross-Cutting Examiner Alert — Unit Conversions

Forgetting to convert cm³ → dm³ gives an answer 1000× too large or too small. This appears in examiner reports across multiple years. Write the conversion as a labelled step — that one line earns its own mark and protects every line after it.

Worked Example 2

2

Calculate Concentration in g/dm³

50 cm³ of a sodium hydroxide solution contains 2.0 g of NaOH. Calculate the concentration in g/dm³. [3 marks]

Step 1 — Convert volume: V = 50 ÷ 1000 = 0.05 dm³

Step 2 — Apply formula: c = mass ÷ volume

Step 3 — Substitute: c = 2.0 ÷ 0.05

c = 40 g/dm³

Mark-Scheme Insurance

Three marks: 1 for the conversion step, 1 for the correct formula, 1 for the final value with unit. Skipping the conversion line costs the first mark even if you got the answer right by doing it in your head.

Relating Concentration, Mass & Volume HT

The HT specification asks you to explain how concentration is related to mass of solute and volume of solution. Two simple proportionalities:

Mass ↑ → Concentration ↑

If you keep the volume the same and increase the mass of solute, the concentration increases proportionally. Doubling the mass doubles the concentration.

Volume ↑ → Concentration ↓

If you keep the mass the same and increase the volume of the solution, the concentration decreases (the same solute is spread over more solution). Doubling the volume halves the concentration.

Rearranging the Formula

The same formula gives you three calculations, depending on what you're solving for:

c = m ÷ V | m = c × V | V = m ÷ c

Volume must be in dm³ in every version of the formula.

Worked Example 3

3

Calculate Mass from Concentration

A solution has a concentration of 25 g/dm³. Calculate the mass of solute in 200 cm³ of this solution. [3 marks]

Step 1 — Convert volume: V = 200 ÷ 1000 = 0.20 dm³

Step 2 — Rearrange formula: mass = concentration × volume

Step 3 — Substitute: m = 25 × 0.20

mass = 5.0 g

Sanity Check

200 cm³ is one-fifth of 1 dm³. One-fifth of 25 g is 5 g. The conversion-then-multiplication method gives the same answer — but the labelled steps are what earn the marks.

Common Errors at a Glance

TopicLoses the markEarns the mark

Limiting reactant identification"The reactants were used up.""The hydrochloric acid was the limiting reactant."

Excess in RP1 (Q04.2)"To use up all the acid." (true of both options — doesn't discriminate)"Excess (insoluble) reactant can be filtered off / removed."

cm³ → dm³ conversionc = 2.0 ÷ 50 = 0.04 g/dm³V = 50 ÷ 1000 = 0.05; c = 2.0 ÷ 0.05 = 40 g/dm³

Mole ratio in limiting calcComparing masses directly to decide which is limiting.Convert each to moles first, then apply the mole ratio.

Final answerNo unit; or unit g/cm³ when working in dm³.Quote unit as g/dm³ for concentration; g for mass.

PEE Framework — Pure Salt from RP1

Question: In RP1, excess zinc oxide is added to dilute hydrochloric acid. Explain how this method results in a pure zinc chloride solution.3 marks

Note: this is different from Q04.2's specific phrasing. Here the question explicitly asks about purity, which is what makes "all the acid reacts" relevant.

P — PointExcess zinc oxide ensures all of the hydrochloric acid is used up, so no acid remains in the final solution.

E — EvidenceZinc oxide is insoluble in water, so the excess does not dissolve into the solution and stays in the mixture as solid particles.

E — ExplanationThe excess solid is then removed by filtration, so the filtrate contains only zinc chloride dissolved in water — a pure solution.

Mark Scheme Sketch

M1 = all acid reacts / no acid in product; M2 = excess solid is insoluble (or stays as a solid); M3 = excess removed by filtration. Compare to Q04.2: when the question only asks why insoluble excess rather than excess acid, only the filtration mark is available — "all the acid reacts" is true of either option.

Practice Question 1

A student dissolves 4.0 g of potassium chloride (KCl) in water and makes the solution up to a total volume of 250 cm³.

Calculate the concentration of the solution in g/dm³. 3 marks

P

Use the formula: c = m ÷ V, where V must be in dm³.

E

Convert volume: V = 250 ÷ 1000 = 0.25 dm³.

E

Substitute and solve: c = 4.0 ÷ 0.25 = 16 g/dm³.

•

Final answer: 16 g/dm³

Practice Question 2 HT

4.8 g of magnesium reacts with 7.3 g of hydrochloric acid in the reaction:

Mg + 2HCl → MgCl₂ + H₂

A_r(Mg) = 24, A_r(H) = 1, A_r(Cl) = 35.5

Identify the limiting reactant and calculate the mass of MgCl₂ produced. 5 marks

P

Calculate moles of each reactant; compare to the mole ratio in the equation.

E

Moles of Mg = 4.8 ÷ 24 = 0.20 mol

E

M_r(HCl) = 1 + 35.5 = 36.5; moles of HCl = 7.3 ÷ 36.5 = 0.20 mol

E

Mole ratio Mg : HCl = 1 : 2. To use up all 0.20 mol Mg would need 0.40 mol HCl — we only have 0.20 mol HCl → HCl is the limiting reactant.

E

Moles of MgCl₂ = moles of HCl ÷ 2 = 0.20 ÷ 2 = 0.10 mol

E

M_r(MgCl₂) = 24 + 2(35.5) = 95; mass = 0.10 × 95 = 9.5 g

•

Final answer: HCl is limiting. Mass of MgCl₂ = 9.5 g.

Summary

Limiting Reactants HT

The reactant used up first determines the maximum product. To find it: calculate moles of each, compare to the equation's mole ratio, name the one that runs out. Always name the substance — "the acid was limiting", not "the reactants ran out".

Concentration of Solutions

c (g/dm³) = m (g) ÷ V (dm³). Convert cm³ → dm³ by dividing by 1000 — write this as a labelled step. Rearrange to find m or V as needed. HT Concentration is proportional to mass and inversely proportional to volume.

Excess in RP1 (Q04.2)

1 mark for: excess (insoluble) reactant can be filtered off / removed. "To use up the acid" is ignored — true of both options.

Most Costly Error

Forgetting cm³ → dm³ conversion → answer 1000× wrong. Always write the conversion as its own line.

Exam Tip

Label every step. "V = 50 ÷ 1000 = 0.05 dm³"; "c = m ÷ V"; "c = 2.0 ÷ 0.05 = 40 g/dm³". For limiting reactants, name the chemical — never "it" or "the reactants".