Chemistry • Year 11 • Module 2 • Lesson 16

Stoichiometry in Solution

Apply the full solution-stoichiometry pathway to real experimental data, compare reactant types, and reason about excess reagent situations.

Apply · Data & Reasoning

1. Interpret experimental data — precipitation reactions in the laboratory

A student performed four precipitation reactions. The table below records the solution data for each. 10 marks

Experiment	Reactant A	V(A) (mL)	c(A) (mol/L)	Precipitate formed
1	AgNO₃	50.0	0.200	AgCl (1:1 ratio with AgNO₃) MM = 143.32 g/mol
2	Pb(NO₃)₂	30.0	0.150	PbI₂ (1:1 ratio with Pb(NO₃)₂) MM = 461.0 g/mol
3	Na₂SO₄	40.0	0.100	BaSO₄ (1:1 ratio with Na₂SO₄) MM = 233.39 g/mol
4	Ca(NO₃)₂	25.0	0.300	CaF₂ (1:1 ratio with Ca(NO₃)₂) MM = 78.07 g/mol

1.1 Complete the “n(A) mol” and “m(precipitate) g” columns. Show the key calculation steps for experiments 1 and 3 in the space below. 8 marks (2 per row: 1 for n, 1 for m)

1.2 Experiment 2 uses a smaller volume but produces more precipitate than Experiment 3. Identify the reason using the data. 2 marks

Stuck? Revisit Worked Example 1 and the drill questions in the lesson. Remember: V in litres first, then n = cV, then m = n × MM.

2. Interpret graph — mass of BaSO₄ precipitate vs concentration of Na₂SO₄

A student reacted 50.0 mL samples of Na₂SO₄ solution (varying concentration) with excess BaCl₂ solution. The mass of BaSO₄ precipitate was recorded. The balanced equation is: Na₂SO₄(aq) + BaCl₂(aq) → BaSO₄(s) + 2NaCl(aq). MM(BaSO₄) = 233.39 g/mol. 8 marks

Figure 2.1. Mass of BaSO₄(s) precipitate produced from 50.0 mL of Na₂SO₄(aq) at varying concentrations, with excess BaCl₂. Equation: Na₂SO₄(aq) + BaCl₂(aq) → BaSO₄(s) + 2NaCl(aq). Illustrative data.

2.1 Describe the relationship between the concentration of Na₂SO₄ and the mass of BaSO₄ precipitate. 2 marks

2.2 Using the graph, estimate the mass of BaSO₄ produced when c(Na₂SO₄) = 0.120 mol/L. Show the calculation that confirms this estimate. 3 marks

2.3 Explain why the graph passes through the origin (0, 0) and what would happen to the slope of the line if the volume of Na₂SO₄ solution used was increased from 50.0 mL to 100.0 mL. 3 marks

Stuck? Revisit Worked Example 1 and the Full Pathway. n(Na₂SO₄) = c × 0.0500 L; ratio 1:1; m(BaSO₄) = n × 233.39.

3. Compare “find mass of product” vs “find concentration of product”

Both question types use the same early steps. Complete the two-column table to show where the pathways diverge. 6 marks (1 per row)

Feature	Find mass of precipitate	Find concentration of product
What you need to know
Step after finding n(product)
Formula used in final step
Volume used in final step
Units of final answer
NSW HSC question clue word

Stuck? Revisit the “Final Answer Types” box in Copy Into Your Books and the two Worked Examples in the lesson.

4. Predict and justify — a water treatment scenario

A water treatment plant near Sydney uses aluminium sulfate, Al₂(SO₄)₃, to remove fluoride ions from drinking water via precipitation. The relevant reaction is:

Al₂(SO₄)₃(aq) + 6NaF(aq) → 2AlF₃(s) + 3Na₂SO₄(aq)

In a test trial, the plant uses 200.0 mL of 0.0500 mol/L Al₂(SO₄)₃ with excess NaF solution. MM(AlF₃) = 83.98 g/mol. 6 marks

4.1 Predict the mass of AlF₃ precipitate formed. Show all steps including the unit conversion. 3 marks

4.2 After the reaction, the total solution volume is 350.0 mL. Calculate the concentration of Na₂SO₄ in the final solution. 3 marks

Stuck? For 4.1: n(Al₂(SO₄)₃) = 0.0500 × 0.200 L. Ratio Al₂(SO₄)₃:AlF₃ = 1:2. For 4.2: ratio Al₂(SO₄)₃:Na₂SO₄ = 1:3; c = n ÷ 0.350 L.

Answers — Do not peek before attempting

Q1.1 — Complete the data table

Exp 1: n(AgNO₃) = 0.200 × 0.0500 = 0.01000 mol; m(AgCl) = 0.01000 × 143.32 = 1.43 g.

Exp 2: n(Pb(NO₃)₂) = 0.150 × 0.0300 = 0.00450 mol; m(PbI₂) = 0.00450 × 461.0 = 2.07 g.

Exp 3: n(Na₂SO₄) = 0.100 × 0.0400 = 0.00400 mol; m(BaSO₄) = 0.00400 × 233.39 = 0.934 g.

Exp 4: n(Ca(NO₃)₂) = 0.300 × 0.0250 = 0.00750 mol; m(CaF₂) = 0.00750 × 78.07 = 0.586 g.

Q1.2 — Why Exp 2 produces more precipitate than Exp 3

Although Exp 2 uses a smaller volume (30.0 mL vs 40.0 mL), its n(Pb(NO₃)₂) = 0.00450 mol exceeds n(Na₂SO₄) = 0.00400 mol. More importantly, the molar mass of PbI₂ (461.0 g/mol) is far greater than that of BaSO₄ (233.39 g/mol), so even though moles are similar, the mass of PbI₂ produced is larger. Award 1 mark for identifying that more moles (or higher c × V product) leads to more precipitate; 1 mark for referencing the difference in molar mass.

Q2.1 — Relationship description

There is a positive linear (directly proportional) relationship between the concentration of Na₂SO₄ and the mass of BaSO₄ precipitate. As concentration doubles, mass of precipitate doubles, consistent with m = n × MM = (c × V) × MM where V and MM are constant.

Q2.2 — Mass at c = 0.120 mol/L

Graph estimate: approximately 1.40 g. Calculation: n(Na₂SO₄) = 0.120 × 0.0500 = 0.00600 mol; ratio 1:1; n(BaSO₄) = 0.00600 mol; m(BaSO₄) = 0.00600 × 233.39 = 1.40 g. Award 1 mark for graph estimate, 1 mark for correct n calculation, 1 mark for correct m calculation.

Q2.3 — Origin and effect of doubling volume

The graph passes through the origin because when c = 0 mol/L, n = 0 and therefore m = 0 — no Na₂SO₄ means no BaSO₄ can form [1]. If volume is increased to 100.0 mL, the slope would double: the same concentration would now produce twice as many moles (n = c × 0.100) and therefore twice the mass of precipitate, giving a steeper line with the same linear shape [2].

Q3 — Compare and contrast table

What you need to know: Mass → molar mass of the precipitate (MM). Concentration → total volume of the final mixed solution.

Step after finding n(product): Mass → multiply by MM. Concentration → divide by total volume in litres.

Formula in final step: Mass → m = n × MM. Concentration → c = n ÷ V(total).

Volume used in final step: Mass → no volume needed. Concentration → V₁ + V₂ in litres.

Units of final answer: Mass → grams (g). Concentration → mol/L (mol L⁻¹).

HSC question clue word: Mass → “what mass forms?” / “calculate the mass of precipitate”. Concentration → “what is the concentration in the final solution?”

Q4.1 — Mass of AlF₃ formed

V = 200.0 mL = 0.2000 L; n(Al₂(SO₄)₃) = 0.0500 × 0.2000 = 0.01000 mol [1].

Ratio Al₂(SO₄)₃ : AlF₃ = 1:2; n(AlF₃) = 0.01000 × 2 = 0.02000 mol [1].

m(AlF₃) = 0.02000 × 83.98 = 1.68 g [1].

Q4.2 — Concentration of Na₂SO₄ in final solution

Ratio Al₂(SO₄)₃ : Na₂SO₄ = 1:3; n(Na₂SO₄) = 0.01000 × 3 = 0.03000 mol [1].

V(total) = 350.0 mL = 0.3500 L [1].

c(Na₂SO₄) = 0.03000 ÷ 0.3500 = 0.0857 mol/L [1].