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.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
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?
The Complete Pathway
n(wanted) = n(given) × ratio (mole ratio, IQ1)
m = n × MM (moles → mass, IQ2)
c = n ÷ V (moles → concentration, IQ3)
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
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
Skills
- Find mass of product from volume + concentration of reactant
- Find concentration of product from stoichiometric data
- Determine which solution reactant is in excess
Concentration
c = n/V
n = cV
V = n/c
Stoichiometry
4-step method
Mole ratios
Balanced equation
Solution Stoichiometry
Most common HSC exam calculation type
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 ÷ Vtotal. 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.
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
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)
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)
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?
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).
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)
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.