Year 11 Chemistry Module 2 IQ3 + IQ1 Merge ⏱ ~40 min Lesson 16 of 20

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.

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📐

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.

Know

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

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

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

Printable worksheet

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Think First

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?

Type your initial thoughts below:

Record your ideas in your workbook.

✏️ Record your ideas in your workbook

📚 Core Content

Key Terms

MoleThe SI unit for amount of substance; contains exactly 6.022 × 10²³ particles.

Avogadro's Number6.022 × 10²³ — the number of particles in one mole of a substance.

Molar MassThe mass of one mole of a substance, measured in g/mol.

Limiting ReagentThe reactant that is completely consumed first, limiting the amount of product formed.

Empirical FormulaThe simplest whole-number ratio of atoms in a compound.

Molecular FormulaThe actual number of atoms of each element in a molecule of a compound.

Misconceptions to Fix

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Wrong: The limiting reagent is the one present in the smallest mass.

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Right: The limiting reagent is the reactant that runs out first based on mole ratios, not mass.

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Where IQ3 Meets IQ1

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.

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.

🧮 Worked Examples

Worked Example 1 — Find mass of precipitate from solution volumes

V + c → mass

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)

1
Step 1 — n(AgNO₃) from solution
V = 30.0 mL = 0.0300 L
n(AgNO₃) = c × V = 0.200 × 0.0300 = 0.00600 mol
2
Step 2 — Mole ratio AgNO₃:AgCl = 1:1
n(AgCl) = 0.00600 mol
3
Step 3 — m(AgCl)
MM(AgCl) = 107.87 + 35.453 = 143.32 g mol⁻¹
m(AgCl) = 0.00600 × 143.32 = 0.860 g

✓ Answer0.860 g of AgCl precipitate

Worked Example 2 — Find concentration of product formed

Solution → c(product)

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)

1
Step 1 — n(HCl)
n(HCl) = 0.100 × 0.0250 = 0.00250 mol
2
Step 2 — Ratio HCl:NaCl = 1:1
n(NaCl) = 0.00250 mol
3
Step 3 — c(NaCl) in 50.0 mL total solution
V(total) = 50.0 mL = 0.0500 L
c(NaCl) = n ÷ V = 0.00250 ÷ 0.0500 = 0.0500 mol/L

✓ Answerc(NaCl) = 0.0500 mol/L

Worked Example 3 — Determine which reagent is in excess

Excess in Solution

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?

1
Calculate n for each reactant
n(Na₂CO₃) = 0.200 × 0.0250 = 0.00500 mol
n(HCl) = 0.100 × 0.0500 = 0.00500 mol
2
Compare to equation ratio: Na₂CO₃:HCl = 1:2
Na₂CO₃ requires 2 × 0.00500 = 0.01000 mol HCl to fully react
Available HCl = 0.00500 mol → NOT enough HCl
3
(a) HCl is the limiting reagent → Na₂CO₃ is in excess
4
(b) How much Na₂CO₃ was consumed?
n(Na₂CO₃ consumed) = n(HCl) × (1÷2) = 0.00500 × 0.500 = 0.00250 mol
n(Na₂CO₃ remaining) = 0.00500 − 0.00250 = 0.00250 mol

✓ Answer(a) Na₂CO₃ is in excess | (b) 0.00250 mol Na₂CO₃ remaining

⚠️

Common Mistakes

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).

📓 Copy Into Your Books

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🔗 Full Pathway

Step 1: n = c × V (convert mL to L first)
Step 2: Write/check balanced equation
Step 3: n(wanted) = n(given) × ratio
Step 4: m = n × MM or c = n ÷ V

⚠️ Volume Rule

ALWAYS convert mL → L before n = cV
Calculate each reactant's n separately
Use total volume only for product concentration
V(total) = V₁ + V₂ when solutions are mixed

🔍 Excess in Solution

n₁ = c₁V₁ and n₂ = c₂V₂ (separately)
Compare n ÷ coefficient for each
Larger value = excess reagent
Calculate remaining: n(given) − n(consumed)

📊 Final Answer Types

"What mass forms?" → m = n × MM
"What concentration results?" → c = n ÷ V(total)
"What volume needed?" → V = n ÷ c
Read the question to decide which to use

🧪 Activities

📊 Activity 1 — Solution Stoichiometry Drill

Volume × Concentration → Moles → Product

Apply the full pathway. Convert mL to L first on every problem.

1 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)
n(AgNO₃) = 0.150 × 0.0500 = 0.00750 mol; ratio 1:1MM(AgCl) = 143.32; m = 0.00750 × 143.32 = 1.07 g
2 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)
n(HCl) = 0.500 × 0.0200 = 0.01000 mol; ratio HCl:CaCl₂ = 2:1n(CaCl₂) = 0.01000 ÷ 2 = 0.00500 molMM(CaCl₂) = 110.98; m = 0.00500 × 110.98 = 0.555 g
3 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)
n(Na₂SO₄) = 0.200 × 0.0300 = 0.00600 mol; ratio 1:1MM(BaSO₄) = 233.39; m = 0.00600 × 233.39 = 1.40 g
4 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?
n(HCl) = 0.250 × 0.0400 = 0.01000 mol; ratio 1:1; n(NaCl) = 0.01000 molV(total) = 80.0 mL = 0.0800 Lc(NaCl) = 0.01000 ÷ 0.0800 = 0.125 mol/L

Show all steps — convert mL to L first:

Complete in workbook.

✏️ Complete in workbook — show V in litres for each step

📈 Activity 2 — Data Analysis: Titration Stoichiometry

Acid–Base Reactions in Solution

The table below shows results from several acid–base neutralisation experiments. Complete the missing values. All experiments use HCl + NaOH → NaCl + H₂O.

Expt	c(HCl)	V(HCl)	n(HCl)	n(NaOH)	c(NaOH) (in 25.0 mL)
A	0.100 mol/L	22.5 mL	0.00225 mol	0.00225 mol	0.0900 mol/L
B	0.200 mol/L	18.0 mL	?	?	?
C	0.150 mol/L	24.0 mL	?	?	?
D	?	20.0 mL	0.00300 mol	?	?

Complete the table:

Copy table into workbook and complete.

✏️ Copy table and complete in workbook

Interactive: Solution Stoichiometry Stepper

Revisit — Think First

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.

✏️ Write a reflection in your workbook

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

✍️ Short Answer

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Extended Questions

ApplyBand 3

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) 4 MARKS

Show all steps:

Answer in workbook.

✏️ Answer in workbook

AnalyseBand 4

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. 5 MARKS

Show all steps:

Answer in workbook.

✏️ Answer in workbook

AnalyseBand 4

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) 6 MARKS

Show all steps for each part:

Answer in workbook.

✏️ Answer in workbook

EvaluateBand 5

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. 4 MARKS

Write a structured critique and correct calculation:

Answer in workbook.

✏️ Answer in workbook

✅ Comprehensive Answers

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📊 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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Tick when you've finished all activities and checked your answers.

Stoichiometryin Solution

The Complete Pathway

Know

Understand

Can Do

Download this lesson's worksheet

Misconceptions to Fix

Where IQ3 Meets IQ1

Concentration

Stoichiometry

Solution Stoichiometry

The Full Pathway

Identifying Excess Reagent in Solution

Worked Example 1 — Find mass of precipitate from solution volumes

Worked Example 2 — Find concentration of product formed

Worked Example 3 — Determine which reagent is in excess

Common Mistakes

📓 Copy Into Your Books

🔗 Full Pathway

⚠️ Volume Rule

🔍 Excess in Solution

📊 Final Answer Types

Volume × Concentration → Moles → Product

Acid–Base Reactions in Solution

Multiple Choice

Extended Questions

✅ Comprehensive Answers

📊 Activity 2 — Data Table

❓ Multiple Choice

📝 Short Answer Model Answers

Mark lesson as complete

Stoichiometry
in Solution