Chemistry · Year 11 · Module 2 · Lesson 12
HSC Exam Practice
Mass–Mass Stoichiometry
Short answer
1.Short answer
State the four steps of the mass–mass stoichiometry method in order. For each step, write the mathematical formula used.
Explain why equation coefficients cannot be used to convert the mass of one reactant directly to the mass of a product. In your answer, refer to what the coefficients actually represent.
Outline the procedure for a “reverse problem” in stoichiometry. In your answer, clarify how it differs from a forward problem and why the same 4-step method applies.
A student uses 16.00 g mol−1 as the molar mass of O2 in a stoichiometry calculation. Identify the error and state its impact on the calculated product mass.
In the reaction N2(g) + 3H2(g) → 2NH3(g), explain why the mole ratio step is essential when converting moles of H2 to moles of NH3, and state the correct ratio.
Data response — copper–silver reaction
2.Data response — Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
A student places a coil of copper wire into a solution of silver nitrate in a glass beaker. After 30 minutes, they observe that the copper wire has partially dissolved, a deposit of silvery-grey crystals has formed on the wire, and the initially colourless solution has turned pale blue. The reaction is:
Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)
The student started with 6.35 g of copper wire. (Cu = 63.546; Ag = 107.87; N = 14.007; O = 15.999)
(a) Calculate the theoretical mass of silver (Ag) deposited when 6.35 g of Cu reacts completely. Show all four steps. (4 marks)
(b) Calculate the mass of AgNO3 that is consumed in this reaction. (3 marks)
(c) The student weighs the silver deposit and finds it is only 19.4 g, not the value you calculated in part (a). Suggest two reasons why the actual yield might be lower than the theoretical yield. (2 marks)
Extended response
3.Extended response
A student burns 48.0 g of magnesium in oxygen according to the equation 2Mg(s) + O2(g) → 2MgO(s). The student claims: “Since the mass of Mg is 48 g and the coefficient ratio of Mg to MgO is 2:2, I expect the product mass to also be 48 g.” (Mg = 24.305, O = 15.999)
Evaluate this claim. In your response, you must:
- Calculate the correct mass of MgO produced using the 4-step method, showing all working.
- Explain why the student’s reasoning is incorrect, with specific reference to molar masses.
- Determine the mass of O2 consumed and use this to verify conservation of mass.
- State a general principle that describes when product mass will exceed reactant mass in a stoichiometry problem.
In the reaction 4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g), a copper-gold mine in Western Australia processes 24.0 g of iron pyrite (FeS2) per batch. (Fe = 55.845, S = 32.06, O = 15.999)
(a) Calculate the mass of SO2 produced per batch. Show all four steps. (4 marks)
(b) The mine must comply with SO2 emission limits of 20.0 g per batch. Using your answer from (a), assess whether the process meets this limit, and suggest one industrial strategy to reduce SO2 emissions. (2 marks)
Chemistry · Year 11 · Module 2 · Lesson 12
Answer Key & Marking Guidelines
Section 1 · Short answer · 4 marks · Band 3
Sample response. Step 1: Write and balance the equation; extract the mole ratio of given to wanted species. (No formula — prerequisite step.) Step 2: Convert given mass to moles: n(given) = m(given) ÷ MM(given). Step 3: Convert moles of given to moles of wanted using the mole ratio: n(wanted) = n(given) × coeff(wanted) ÷ coeff(given). Step 4: Convert moles of wanted to mass: m(wanted) = n(wanted) × MM(wanted).
Marking notes. 1 mark per step: 1 = Step 1 stated (balance, extract ratio); 1 = Step 2 with formula n = m ÷ MM; 1 = Step 3 with mole ratio formula; 1 = Step 4 with formula m = n × MM. Deduct if formulas are missing or wrong.
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. Equation coefficients compare the numbers of moles of each species, not their masses. Different substances have different molar masses, so equal numbers of moles weigh different amounts. For example, 1 mol of C (12 g) reacts with 1 mol of O2 (32 g) in a 1:1 ratio, but they are not equal in mass. Using coefficients to convert mass directly would give a wrong answer unless all species coincidentally had identical molar masses.
Marking notes. 1 = coefficients compare moles (not masses); 1 = different substances have different molar masses so equal moles ≠ equal mass; 1 = applies this to a specific example or states that mass must always be converted through moles.
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. In a reverse problem, the mass of a product (not a reactant) is given, and the task is to find the mass of a reactant. The 4-step method is applied identically: the “given substance” is the product whose mass is known (Step 2), the mole ratio converts product moles to reactant moles (Step 3), and the reactant mass is found using its molar mass (Step 4). The direction of the problem does not require a different method because the 4-step algorithm applies regardless of which species is given.
Marking notes. 1 = correctly defines reverse problem (given = product, wanted = reactant); 1 = states same 4 steps are used; 1 = explains the “given” and “wanted” roles are simply swapped (product becomes given in Step 2).
Section 1 · Short answer · 2 marks · Band 3
Sample response. The error is using the atomic mass of oxygen (16.00 g mol−1) instead of the molar mass of the diatomic molecule O2 (32.00 g mol−1). This error halves n(O2) in Step 2 (or halves the product mass if O2 is the wanted substance in Step 4), producing an answer that is half the correct value.
Marking notes. 1 = identifies error (using atomic mass 16 instead of molecular mass 32 for O2); 1 = states the impact (answer is half the correct value or n(O2) is doubled/halved incorrectly).
Section 1 · Short answer · 2 marks · Band 4
Sample response. The mole ratio step is essential because H2 and NH3 have different coefficients in the balanced equation (3 and 2, respectively); they are not present in a 1:1 ratio. Every 3 moles of H2 produces only 2 moles of NH3. Skipping the ratio step would assume n(NH3) = n(H2), giving an answer 50% higher than correct. The correct ratio is H2 : NH3 = 3 : 2, so n(NH3) = n(H2) × 2 ÷ 3.
Marking notes. 1 = states that H2:NH3 is not 1:1 and identifies the correct ratio 3:2; 1 = explains the impact of skipping the ratio (overcounting moles of NH3).
Section 2 · Data response · 9 marks · Band 4–5
Part (a) — mass of Ag (4 marks). Step 1: Balanced. Ratio Cu : Ag = 1 : 2 [1]. Step 2: n(Cu) = 6.35 ÷ 63.546 = 0.09992 mol [1]. Step 3: n(Ag) = 0.09992 × 2 ÷ 1 = 0.1998 mol [1]. Step 4: m(Ag) = 0.1998 × 107.87 = 21.6 g [1].
Part (b) — mass of AgNO3 (3 marks). Ratio Cu : AgNO3 = 1 : 2 [1]. n(AgNO3) = 0.09992 × 2 = 0.1998 mol [1]. MM(AgNO3) = 107.87 + 14.007 + 3(15.999) = 169.88; m(AgNO3) = 0.1998 × 169.88 = 33.9 g [1].
Part (c) — reasons for lower yield (2 marks). Any two valid reasons: (1) Reaction was incomplete — not all copper reacted (limited reaction time or insufficient AgNO3). (2) Some silver crystals remained in solution or fell off the wire and were not collected on weighing. (3) The copper wire contained impurities that did not react. (4) Experimental error in measuring initial copper mass. Award 1 mark per valid distinct reason.
Section 3 · Extended response · 8 marks · Band 5–6
4-step calculation for MgO (4 marks). Step 1: Ratio Mg : MgO = 2 : 2 = 1 : 1 [1]. Step 2: MM(Mg) = 24.305; n(Mg) = 48.0 ÷ 24.305 = 1.976 mol [1]. Step 3: n(MgO) = 1.976 × 1 = 1.976 mol [1]. Step 4: MM(MgO) = 24.305 + 15.999 = 40.304; m(MgO) = 1.976 × 40.304 = 79.6 g [1]. (Accept 79.5–79.7 g.)
Why student’s reasoning is incorrect (1 mark). The student confuses equal moles with equal mass. The ratio 2:2 = 1:1 means equal moles of Mg and MgO, but Mg has MM = 24.305 and MgO has MM = 40.304. Equal moles of a heavier substance must weigh more. Each oxygen atom (MM ≈ 16.0) is incorporated from O2 into every MgO unit, adding 16 g per mole of MgO.
O2 consumed and conservation verification (2 marks). Ratio Mg : O2 = 2 : 1. n(O2) = 1.976 × 1 ÷ 2 = 0.9880 mol; m(O2) = 0.9880 × 31.998 = 31.6 g [1]. Total reactants = 48.0 + 31.6 = 79.6 g = m(MgO) ⇒ mass is conserved [1].
General principle (1 mark). Product mass exceeds reactant (given) mass whenever the wanted substance incorporates atoms from other reactants (e.g. oxygen), so its molar mass is higher than the given substance. Product mass can be less than reactant mass if the product has a lower molar mass. Mass is always conserved across all species, but the mass of any single product need not equal the mass of any single reactant.
Marking criteria summary (8 marks): 1 = correct ratio; 1 = n(Mg); 1 = n(MgO); 1 = m(MgO) = 79.6 g; 1 = explains error in student’s reasoning (equal moles ≠ equal mass); 1 = m(O2) calculated; 1 = conservation verified (total reactants = total products); 1 = general principle stated.
Section 3 · Extended response · 6 marks · Band 4–5
Part (a) — mass of SO2 (4 marks). Step 1: Ratio FeS2 : SO2 = 4 : 8 = 1 : 2 [1]. Step 2: MM(FeS2) = 55.845 + 2(32.06) = 119.97; n(FeS2) = 24.0 ÷ 119.97 = 0.2001 mol [1]. Step 3: n(SO2) = 0.2001 × 2 ÷ 1 = 0.4002 mol [1]. Step 4: MM(SO2) = 32.06 + 2(15.999) = 64.058; m(SO2) = 0.4002 × 64.058 = 25.6 g [1].
Part (b) — emission assessment and strategy (2 marks). The process produces 25.6 g of SO2 per batch, which exceeds the 20.0 g limit by 5.6 g. The process does not comply [1]. One valid industrial strategy: (i) scrub exhaust gases with limestone (CaCO3) or lime (CaO) to absorb SO2 as CaSO3/CaSO4 (flue-gas desulfurisation); or (ii) reduce the mass of FeS2 processed per batch to below the threshold (stoichiometric calculation: max mass = 20.0 ÷ 25.6 × 24.0 = 18.75 g per batch); or (iii) convert SO2 to sulfuric acid (H2SO4) via the Contact process to capture the gas as a saleable product. Award 1 mark for any scientifically valid, named strategy [1].