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Module 2 · L16 of 20 40 min ⚡ +50 XP in Learn · +25 to complete IQ3 + IQ1 Merge

Stoichiometry in Solution

From Lesson 6 you know how to find moles from a solution's concentration and volume. From Lesson 11 you know how to use mole ratios from balanced equations. This lesson fuses both skills — it's the most powerful calculation type in the module, and one of the most common in HSC exams.

Today's hook — From Lesson 6 you know how to find moles from a solution's concentration and volume. From Lesson 11 you know how to use mole ratios from balanced equations. This lesson fuses both skills — it's the most powerful calculation type in the module, and one of the most common in HSC exams.

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Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Recall — your gut answer first

+5 XP warm-up

You mix 25 mL of silver nitrate solution (AgNO₃) with excess sodium chloride solution (NaCl) and a white precipitate forms: AgNO₃ + NaCl → AgCl↓ + NaNO₃. If you only know the volume and concentration of the AgNO₃ solution, what is the minimum set of steps you need to find the mass of AgCl precipitate — and where does stoichiometry fit in?

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Formula reference · this lesson

core formula

📐

The Complete Pathway

n = c × V     (concentration → moles, IQ3)
n(wanted) = n(given) × ratio     (mole ratio, IQ1)
m = n × MM     (moles → mass, IQ2)
c = n ÷ V     (moles → concentration, IQ3)

⚠️ Volume must ALWAYS be in litres before using n = c × V. 25 mL = 0.025 L. Forgetting this conversion is the single most common error in solution stoichiometry.

What you'll master

Know

Key facts

n = cV links concentration to moles
Solution stoichiometry combines IQ3 and IQ1 skills
mL → L conversion is non-negotiable first step
Full pathway: V × c → n → ratio → n → m or c

Understand

Concepts

Why concentration × volume gives moles (not grams)
When to find m vs when to find c as the final answer
How to determine excess reagent in a solution context

Can do

Skills

Find mass of product from volume + concentration of reactant
Find concentration of product from stoichiometric data
Determine which solution reactant is in excess

Key terms

n = c × V

Moles = concentration (mol L⁻¹) × volume (L); fundamental equation for solution stoichiometry.

Solution stoichiometry

Calculations using moles derived from concentration and volume of reacting solutions.

Mole ratio

The ratio of stoichiometric coefficients from the balanced equation; used to relate moles of reactant to moles of product.

Titration stoichiometry

n(acid) × stoich. coefficient ratio = n(base) at equivalence point; used to find unknown concentration.

Dilution calculation

c₁V₁ = c₂V₂; used when a solution is diluted; moles of solute remain constant.

Units consistency

Volume must be in litres for c × V = n; concentration in mol L⁻¹; convert mL to L by dividing by 1000.

Where IQ3 Meets IQ1

core concept

Inquiry Question 3

Concentration

c = n/V
n = cV
V = n/c

Inquiry Question 1

Stoichiometry

4-step method
Mole ratios
Balanced equation

This Lesson

Solution Stoichiometry

Most common HSC exam calculation type

The Full Pathway

The only new skill is combining n = cV with the existing 4-step method. You already know all the pieces — this lesson shows you how to chain them.

Which answer type does the question want?
If the question asks "what mass of precipitate forms?" → convert moles to mass (m = n × MM).
If the question asks "what is the concentration of the product?" → divide moles by volume (c = n ÷ V).
If the question asks "what volume contains a given amount?" → rearrange (V = n ÷ c).
Read the question carefully — this determines your final step.

Solution stoichiometry combines n = c × V (from IQ3) with the 4-step mole-ratio method (from IQ1). Full pathway: n(A) = c(A) × V(A) → apply mole ratio → n(B) → m(B) = n × MM or c(B) = n ÷ V_total. Always convert mL to L before substituting; each reactant's moles come from its own c and V independently.

Pause — copy the highlighted pathway into your book before moving on.

Identifying Excess Reagent in Solution

When two solutions are mixed, one reactant may be in excess. The method is identical to the limiting reagent method from L13 — convert both to moles, compare to the equation ratio, identify which runs out first. Then calculate based on the limiting reagent.

Tip for "which is in excess?" questions: Calculate the moles of each reactant from c × V. Apply the ratio comparison. The one with the larger (n ÷ coefficient) value is in excess. You can then calculate the moles of excess remaining after reaction.

Quick check: 25.0 mL of 0.200 mol/L AgNO₃ is mixed with excess NaCl. Which is the correct first step to find the mass of AgCl precipitate?

Worked examples · reveal as you go

Worked Example 1 — Find mass of precipitate from solution volumes +5 XP on full reveal

30.0 mL of 0.200 mol/L AgNO₃ solution is mixed with excess NaCl solution. A white AgCl precipitate forms. Calculate the mass of AgCl precipitate. AgNO₃ + NaCl → AgCl + NaNO₃. (Ag = 107.87, Cl = 35.453)

V = 30.0 mL = 0.0300 L

n(AgNO₃) = c × V = 0.200 × 0.0300 = 0.00600 mol

Step 1 — n(AgNO₃) from solution

n(AgCl) = 0.00600 mol

Step 2 — Mole ratio AgNO₃:AgCl = 1:1

MM(AgCl) = 107.87 + 35.453 = 143.32 g mol⁻¹

m(AgCl) = 0.00600 × 143.32 = 0.860 g

Step 3 — m(AgCl)

Worked Example 2 — Find concentration of product formed +5 XP on full reveal

25.0 mL of 0.100 mol/L HCl is neutralised by 25.0 mL of NaOH solution: HCl + NaOH → NaCl + H₂O. The resulting NaCl solution occupies 50.0 mL total. Find the concentration of NaCl in the final solution. (Na = 22.990, Cl = 35.453)

n(HCl) = 0.100 × 0.0250 = 0.00250 mol

Step 1 — n(HCl)

n(NaCl) = 0.00250 mol

Step 2 — Ratio HCl:NaCl = 1:1

V(total) = 50.0 mL = 0.0500 L

c(NaCl) = n ÷ V = 0.00250 ÷ 0.0500 = 0.0500 mol/L

Step 3 — c(NaCl) in 50.0 mL total solution

Worked Example 3 — Determine which reagent is in excess +5 XP on full reveal

25.0 mL of 0.200 mol/L Na₂CO₃ is mixed with 50.0 mL of 0.100 mol/L HCl: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (a) Which reactant is in excess? (b) How many moles of excess remain after the reaction?

n(Na₂CO₃) = 0.200 × 0.0250 = 0.00500 mol

n(HCl) = 0.100 × 0.0500 = 0.00500 mol

Calculate n for each reactant

Na₂CO₃ requires 2 × 0.00500 = 0.01000 mol HCl to fully react

Available HCl = 0.00500 mol → NOT enough HCl

Compare to equation ratio: Na₂CO₃:HCl = 1:2

(a) HCl is the limiting reagent → Na₂CO₃ is in excess

n(Na₂CO₃ consumed) = n(HCl) × (1÷2) = 0.00500 × 0.500 = 0.00250 mol

n(Na₂CO₃ remaining) = 0.00500 − 0.00250 = 0.00250 mol

(b) How much Na₂CO₃ was consumed?

Sort the steps+7 XP

Click two steps to swap them. Order the solution-stoichiometry method to solve: 30.0 mL of 0.200 mol/L AgNO₃ is mixed with excess NaCl. What mass of AgCl precipitate forms? (AgNO₃ + NaCl → AgCl + NaNO₃; MM(AgCl) = 143.32)

Apply the mole ratio AgNO₃ : AgCl = 1 : 1, so n(AgCl) = 0.00600 mol.
Convert the volume to litres: V = 30.0 mL = 0.0300 L.
Final answer: m(AgCl) = 0.860 g.
Calculate moles of the given species: n(AgNO₃) = c × V = 0.200 × 0.0300 = 0.00600 mol.
Convert moles of product to mass: m(AgCl) = n × MM = 0.00600 × 143.32.

Common errors · the 3 traps that cost marks

Plugging mL into n = cV instead of litres

n = c × V only works when V is in litres. Using 25.0 instead of 0.0250 gives an answer 1000× too large. The answer may still look plausible (especially for small concentrations), making this error hard to catch unless you check units.

✓ Fix: The very first thing you write after reading a solution stoichiometry problem should be: "V = ___ mL = ___ L". Make this conversion visible before any calculation.

Using the total combined volume for calculating n instead of the individual solution volumes

When two solutions are mixed (e.g. 25 mL AgNO₃ + 50 mL NaCl), each reactant's moles must be calculated from its own volume and concentration separately. Using the total volume (75 mL) for either reactant gives a wrong n value.

✓ Fix: Calculate n for each reactant independently: n₁ = c₁ × V₁ and n₂ = c₂ × V₂. Only use total volume if you're calculating the concentration of a product in the final mixed solution.

Forgetting that total volume changes when solutions are mixed

When finding the concentration of a product in a mixed solution, you must use the total final volume — not the volume of one of the original solutions. If 30 mL of A is mixed with 20 mL of B, the total volume for the product concentration calculation is 50 mL = 0.050 L.

✓ Fix: When finding c(product), always calculate: V(total) = V₁ + V₂. Then c = n ÷ V(total).

Work mode · how are you completing this lesson?

Quick-fire practice · 5 reps +2 XP per reveal

AgNO₃ + NaCl → AgCl↓ + NaNO₃
50.0 mL of 0.150 mol/L AgNO₃ reacts with excess NaCl. Find the mass of AgCl precipitate. (Ag = 107.87, Cl = 35.453)

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
20.0 mL of 0.500 mol/L HCl reacts with excess Ca(OH)₂. Find the mass of CaCl₂ produced. (Ca = 40.078, Cl = 35.453)

Na₂SO₄ + BaCl₂ → BaSO₄↓ + 2NaCl
30.0 mL of 0.200 mol/L Na₂SO₄ reacts with excess BaCl₂. What mass of BaSO₄ precipitate forms? (Ba = 137.33, S = 32.06, O = 15.999)

HCl + NaOH → NaCl + H₂O
40.0 mL of 0.250 mol/L HCl is mixed with 40.0 mL of 0.250 mol/L NaOH. What is the concentration of NaCl in the final 80.0 mL solution?

What mass of NaOH is needed to prepare 250 mL of 0.20 mol L⁻¹ solution? (M(NaOH) = 40.0)

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Revisit your thinking

At the start of this lesson, you thought about the steps needed to find the mass of AgCl precipitate from the volume and concentration of an AgNO₃ solution.

The full pathway is: (1) convert volume to litres; (2) calculate n(AgNO₃) = c × V; (3) apply the mole ratio from the balanced equation; (4) convert moles to mass using m = n × MM. When two solutions are mixed, the product dissolves in the total combined volume — not just one of the original volumes. Calculate moles from each solution separately, then apply stoichiometry. Use the total combined volume only when finding the final concentration of the product.

Reflect: how did your initial thinking compare to what you've learned?

Write a reflection in your workbook.

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Interactive Tool — Stoichiometry Calculator Open fullscreen ↗

Use the Stoichiometry Calculator. How many moles are in 44 g of CO₂ (molar mass = 44 g/mol)?

Multiple choice

+2 XP per correct · +5 bonus if perfect

Pick your answer, then rate your confidence — that tells the system what to drill next.

Fill the blanks+4 XP

Complete the working below. 25.0 mL of 0.100 mol/L HCl is neutralised by 25.0 mL of NaOH. (HCl + NaOH → NaCl + H₂O.) Find the concentration of NaCl in the final 50.0 mL solution.

Step 1 — Convert volume: V(HCl) = 25.0 mL = L.

Step 2 — Moles of given species: n(HCl) = c × V = 0.100 × 0.0250 = mol.

Step 3 — Apply mole ratio HCl : NaCl = 1 : , so n(NaCl) = 0.00250 mol.

Step 4 — Total volume of the mixed solution = L.

Step 5 — c(NaCl) = n ÷ V(total) = 0.00250 ÷ 0.0500 = mol/L.

Short answer

ApplyBand 34 MARKS

Q1. 6. 30.0 mL of 0.250 mol/L Pb(NO₃)₂ is mixed with excess KI solution: Pb(NO₃)₂ + 2KI → PbI₂↓ + 2KNO₃. (a) Calculate the moles of Pb(NO₃)₂. (b) Calculate the mass of PbI₂ precipitate formed. (Pb = 207.2, I = 126.90)

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AnalyseBand 45 MARKS

Q2. 7. 50.0 mL of 0.200 mol/L HCl is mixed with 50.0 mL of 0.100 mol/L Na₂CO₃: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (a) Identify the limiting reagent, showing your full comparison. (b) Calculate the concentration of NaCl in the final 100 mL solution.

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AnalyseBand 46 MARKS

Q3. 8. 40.0 mL of 0.150 mol/L CuSO₄ is mixed with 60.0 mL of 0.150 mol/L NaOH: CuSO₄ + 2NaOH → Cu(OH)₂↓ + Na₂SO₄. (a) Determine the limiting reagent with full working. (b) Calculate the mass of Cu(OH)₂ precipitate formed. (c) Calculate the concentration of Na₂SO₄ in the final solution. (Cu=63.546, O=15.999, H=1.008, Na=22.990, S=32.06)

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EvaluateBand 54 MARKS

Q4. 9. A student wants to find the concentration of an unknown H₂SO₄ solution by reacting it with a 0.100 mol/L NaOH solution. They use 25.0 mL of H₂SO₄ and add 40.0 mL of NaOH. The equation is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. They calculate c(H₂SO₄) = (0.100 × 0.040) ÷ 0.025 = 0.160 mol/L. Identify and correct any errors in their method.

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📖 Comprehensive answers (click to reveal)

Activity 2 — Data Table

B: n(HCl) = 0.200 × 0.0180 = 0.00360 mol; n(NaOH) = 0.00360 mol; c(NaOH) = 0.00360 ÷ 0.0250 = 0.144 mol/L

C: n(HCl) = 0.150 × 0.0240 = 0.00360 mol; n(NaOH) = 0.00360 mol; c(NaOH) = 0.00360 ÷ 0.0250 = 0.144 mol/L

D: c(HCl) = 0.00300 ÷ 0.0200 = 0.150 mol/L; n(NaOH) = 0.00300 mol; c(NaOH) = 0.00300 ÷ 0.0250 = 0.120 mol/L

❓ Multiple Choice

1. C — n = 0.100 × 0.0250 = 0.00250 mol; ratio 1:1; n(AgCl) = 0.00250 mol.

2. B — n(H₂SO₄) = 0.500 × 0.030 = 0.0150 mol; ratio 1:2; n(NaOH) = 0.0300 mol.

3. A — When two solutions mix, the product is dissolved in the total combined volume.

4. D — Used 25.0 mL directly instead of 0.0250 L, giving an answer 1000× too large.

5. B — n(Na₂CO₃) = 0.200 × 0.020 = 0.00400 mol; requires 0.00800 mol HCl; n(HCl) available = 0.200 × 0.020 = 0.00400 mol < 0.00800 → HCl is limiting.

6. D — n(NaCl) = 0.200 × 0.050 = 0.01000 mol; ratio 1:1; n(NaCl) = 0.01000 mol; V(total) = 50+50 = 100 mL = 0.100 L; c(NaCl) = 0.01000 ÷ 0.100 = 0.100 mol/L. Student B made the error of not accounting for the dilution that occurs when two solutions are combined.

7. B — To calculate n(BaSO₄), you need n(BaCl₂) = c × V. This requires both concentration AND volume. Volume alone (A) gives no information without concentration. Concentration alone (D) gives no information without volume. Volumes of both reagents (C) without concentrations is insufficient.

Short Answer Model Answers

Q6 (4 marks):

(a) V = 30.0 mL = 0.0300 L; n(Pb(NO₃)₂) = 0.250 × 0.0300 = 0.00750 mol (b) Ratio 1:1; n(PbI₂) = 0.00750 mol; MM(PbI₂) = 207.2 + 2(126.90) = 461.0 m(PbI₂) = 0.00750 × 461.0 = 3.46 g

Q7 (5 marks):

(a) n(HCl) = 0.200 × 0.0500 = 0.01000 mol; ÷ 2 = 0.00500 n(Na₂CO₃) = 0.100 × 0.0500 = 0.00500 mol; ÷ 1 = 0.00500

Both give 0.00500 — they are in exact stoichiometric proportion; neither is in excess. Both are consumed completely.

(b) Na₂CO₃:NaCl = 1:2; n(NaCl) = 0.00500 × 2 = 0.01000 mol V(total) = 100 mL = 0.100 L; c(NaCl) = 0.01000 ÷ 0.100 = 0.100 mol/L

Q8 (6 marks):

(a) n(CuSO₄) = 0.150 × 0.040 = 0.00600 mol; ÷ 1 = 0.00600 n(NaOH) = 0.150 × 0.060 = 0.00900 mol; ÷ 2 = 0.00450

NaOH gives the smaller value (0.00450 < 0.00600) → NaOH is the limiting reagent.

(b) Ratio NaOH:Cu(OH)₂ = 2:1; n(Cu(OH)₂) = 0.00900 ÷ 2 = 0.00450 mol MM(Cu(OH)₂) = 63.546 + 2(15.999 + 1.008) = 97.560 g/mol m(Cu(OH)₂) = 0.00450 × 97.560 = 0.439 g (c) Ratio NaOH:Na₂SO₄ = 2:1; n(Na₂SO₄) = 0.00900 ÷ 2 = 0.00450 mol V(total) = 40.0 + 60.0 = 100.0 mL = 0.1000 L; c(Na₂SO₄) = 0.00450 ÷ 0.1000 = 0.0450 mol/L

Q9 (4 marks): Error: the student calculated n(NaOH) correctly (0.00400 mol) but then used it directly as n(H₂SO₄) without applying the mole ratio (2NaOH : 1H₂SO₄). This means they treated the ratio as 1:1 instead of 2:1, giving an answer twice too large.

Correct: n(NaOH) = 0.100 × 0.040 = 0.00400 mol Ratio 2:1 → n(H₂SO₄) = 0.00400 ÷ 2 = 0.00200 mol c(H₂SO₄) = 0.00200 ÷ 0.025 = 0.0800 mol/L

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