HSC Exam Practice

Sample response. The limiting reagent is the reactant that is completely consumed first in a chemical reaction; it determines the maximum amount of product that can form. The excess reagent is the reactant present in greater than stoichiometric amount; some remains unreacted when the reaction stops. One consequence of having a limiting reagent is that the reaction stops when it is consumed, regardless of how much excess reagent remains.

Marking notes. 1 mark for correct definition of limiting reagent (completely consumed, determines maximum yield); 1 mark for correct definition of excess reagent (more than stoichiometric amount, remains unreacted); 1 mark for a correct consequence (reaction stops, product yield is limited, excess reagent left over).

Sample response. Step 1: Ensure the chemical equation is balanced and note the stoichiometric coefficient of each reactant. Step 2: Convert the given mass of each reactant to moles using n = m ÷ MM. Step 3: Divide the moles of each reactant by its stoichiometric coefficient to obtain a ‘reaction-unit’ quotient. Step 4: The reactant with the smaller quotient is the limiting reagent. All theoretical yield calculations then use only the limiting reagent’s moles.

Marking notes. 1 mark for identifying step of confirming balanced equation and noting coefficients; 1 mark for converting mass to moles (n = m ÷ MM); 1 mark for dividing each n by its coefficient; 1 mark for identifying the smaller quotient as the limiting reagent and stating yield is calculated from LR only.

Sample response. The mass of a substance depends on both its molar mass and the number of moles present. A smaller mass might correspond to more moles if the molar mass is low. Furthermore, even if one reactant has more moles, a larger stoichiometric coefficient may require more of that substance to be consumed per ‘unit’ of reaction. For example, in 2Na + Cl₂ → 2NaCl: suppose n(Na) = 0.44 mol and n(Cl₂) = 0.28 mol. Raw mole comparison suggests Cl₂ has fewer moles, so Cl₂ is limiting. However, dividing by coefficients: Na = 0.44 ÷ 2 = 0.22; Cl₂ = 0.28 ÷ 1 = 0.28. Na has the smaller quotient, so Na is actually the limiting reagent — despite having more raw moles than Cl₂.

Marking notes. 1 mark for explaining that molar mass affects how many moles a given mass represents; 1 mark for explaining that the coefficient determines how many moles are consumed per ‘reaction-unit’; 1 mark for a specific numerical or named example demonstrating that the smaller-mass or smaller-mole reactant is not necessarily limiting.

Sample response. Once the limiting reagent (LR) is identified, calculate the moles of the excess reagent that are consumed using the mole ratio from the balanced equation: n(excess consumed) = n(LR) × [coeff(excess) ÷ coeff(LR)]. Then subtract from the original moles: n(excess remaining) = n(excess original) − n(excess consumed). Finally, convert to mass: m(excess remaining) = n(excess remaining) × MM(excess).

Marking notes. 1 mark for formula n(excess consumed) = n(LR) × [coeff(excess) ÷ coeff(LR)]; 1 mark for n(remaining) = n(original) − n(consumed); 1 mark for m = n × MM applied correctly to excess.

Sample response. n(Mg) = 4.86 ÷ 24.305 = 0.2000 mol. n(O₂) = 3.20 ÷ 31.998 = 0.1000 mol. Mg: 0.2000 ÷ 2 = 0.1000. O₂: 0.1000 ÷ 1 = 0.1000. Both quotients are equal (0.1000 = 0.1000). Neither reactant is limiting — the reactants are in exactly stoichiometric proportions. Both will be completely consumed and there is no excess.

Marking notes. 1 mark for both correct n ÷ coeff quotients; 1 mark for correctly stating the quotients are equal; 1 mark for concluding that the reactants are in stoichiometric ratio and neither is limiting (both consumed completely).

Sample response (a) — LR identification (3 marks). n(Fe) = 5.58 ÷ 55.845 = 0.09992 mol; n(Cl₂) = 7.10 ÷ 70.906 = 0.1001 mol. Fe: 0.09992 ÷ 2 = 0.04996. Cl₂: 0.1001 ÷ 3 = 0.03337. Cl₂ has the smaller quotient (0.03337 < 0.04996). Cl₂ is the limiting reagent. [1 mark for both n values; 1 mark for both quotients; 1 mark for correct LR with explicit comparison]

Sample response (b) — Theoretical yield of FeCl₃ (3 marks). LR = Cl₂; n(Cl₂) = 0.1001 mol. Ratio Cl₂:FeCl₃ = 3:2. n(FeCl₃) = 0.1001 × (2÷3) = 0.06673 mol. MM(FeCl₃) = 55.845 + 3(35.453) = 162.20 g/mol. m(FeCl₃) = 0.06673 × 162.20 = 10.8 g. [1 mark for correct n from LR; 1 mark for MM; 1 mark for correct mass]

Sample response (c) — Excess Fe remaining (2 marks). Ratio Cl₂:Fe = 3:2. n(Fe consumed) = 0.1001 × (2÷3) = 0.06673 mol. n(Fe remaining) = 0.09992 − 0.06673 = 0.03319 mol. m(Fe remaining) = 0.03319 × 55.845 = 1.85 g of Fe remains unreacted. [1 mark for correct consumed; 1 mark for correct remaining mass]

Sample response (d) — Evaluating the student’s claim (2 marks). The student’s claim is incorrect. While Fe does have fewer moles (0.09992) than Cl₂ (0.1001), the limiting reagent cannot be determined by comparing raw moles. After dividing by coefficients, Cl₂ has the smaller quotient (0.03337) compared to Fe (0.04996), confirming Cl₂ is the limiting reagent. The student failed to account for the fact that the equation requires 3 moles of Cl₂ for every 2 moles of Fe; Cl₂ is consumed at a faster rate relative to its supply. [1 mark for identifying the error — raw mole comparison; 1 mark for using the actual quotients to explain why Cl₂ is the LR despite having more raw moles]

Sample response. Correctly identifying the limiting reagent is critical in industrial chemical production because it determines the maximum theoretical yield — the upper bound on how much product can be obtained from a given set of reactants. If the limiting reagent is incorrectly identified and the excess reagent’s moles are used to calculate theoretical yield, the yield is overestimated. For example, in the Haber process (N₂ + 3H₂ → 2NH₃), if 28.0 g of N₂ and 9.00 g of H₂ are used: n(N₂) = 1.000 mol (÷ 1 = 1.000); n(H₂) = 4.464 mol (÷ 3 = 1.488). N₂ is the LR, giving m(NH₃) = 34.0 g. If a process engineer mistakenly uses H₂ (the excess reagent), they would calculate n(NH₃) = 4.464 × (2÷3) = 2.976 mol, m = 50.7 g — an overestimate of 16.7 g. An industrial plant designed around the incorrect 50.7 g figure would order excess raw materials, waste feedstock, oversize reactors, and misallocate energy inputs, all increasing production costs. For water treatment plants using FeCl₃, overestimating yield means insufficient product is produced per batch, forcing unplanned production runs that increase labour and energy costs. A chemist might deliberately add one reactant in large excess to drive the reaction towards completion (Le Chatelier’s principle for equilibrium reactions, or to ensure a rare/expensive reactant is fully consumed in a single-step reaction). For example, in synthesising a pharmaceutical intermediate, the cheaper reactant is added in excess to maximise yield of the expensive one. The trade-off is that the excess reagent must be separated and recovered after reaction — adding a purification step. In summary, correct LR identification is the foundation of accurate yield prediction, cost estimation, and efficient resource use in industrial chemistry.

Marking criteria (7 marks). 1 = correctly explains that using the excess reagent’s moles overestimates theoretical yield. 1 = provides a specific numerical example demonstrating the overestimation (e.g. Haber process or FeCl₃). 1 = analyses the industrial consequence of overestimated yield in terms of cost or materials waste. 1 = analyses a second industrial consequence (energy, process efficiency, or under-production). 1 = describes a scenario where deliberate excess is used (drive reaction to completion / protect expensive reagent). 1 = explains the trade-off (recovery/separation cost). 1 = reaches an explicit evaluative conclusion linking correct LR identification to industrial efficiency and product quality, using precise chemical terminology throughout.