Chemistry • Year 11 • Module 2 • Lesson 13
Limiting Reagents & Theoretical Yield
Lock in the core vocabulary, the five-step LR identification method, and the key difference between limiting and excess reagents before tackling harder problems.
1. Term–definition match
The definitions below are shuffled. Write the matching term from this list in the right-hand column: limiting reagent, excess reagent, theoretical yield, stoichiometric ratio, mole, molar mass, n ÷ coefficient comparison, excess remaining. 8 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The reactant that is completely consumed first and determines the maximum amount of product that can form. | |
| 1.2 | The reactant present in greater than the stoichiometric amount; some remains unused when the reaction stops. | |
| 1.3 | The maximum mass of product calculated from the stoichiometry of the limiting reagent, assuming 100% conversion. | |
| 1.4 | The fixed proportion in which reactants combine, given by the coefficients in the balanced equation. | |
| 1.5 | The SI unit of amount of substance, equal to 6.022 × 1023 particles. | |
| 1.6 | The mass in grams of one mole of a substance, in g/mol. | |
| 1.7 | The method used to identify the limiting reagent: divide each reactant's moles by its coefficient and compare the results. | |
| 1.8 | n(excess original) − n(excess consumed); used to find how much of the excess reagent is left over after reaction. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below. 12 marks (1 T/F + 1 correction each)
2.1 The limiting reagent is always the reactant with the smaller mass. T / F
2.2 To identify the limiting reagent, you should convert both reactant masses to moles, then divide each by its stoichiometric coefficient; the reactant with the smaller quotient is the limiting reagent. T / F
2.3 Theoretical yield can be calculated from either the limiting or the excess reagent — you get the same answer either way. T / F
2.4 The excess reagent is the reactant that runs out first in a chemical reaction. T / F
2.5 After the limiting reagent is consumed, the reaction stops even if there is still some excess reagent present. T / F
2.6 In the reaction 2Na + Cl2 → 2NaCl, comparing 0.44 mol Na and 0.28 mol Cl2 by their raw moles tells you directly which is the limiting reagent. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
coefficient · consumed · excess · limiting · molar mass · moles · stoichiometric · theoretical yield
In most reactions, reactants are not mixed in perfect ___________ proportions. The reactant that is completely ___________ first is called the ___________ reagent. Any remaining reactant is said to be in ___________. To identify which reactant is limiting, convert the mass of each reactant to ___________ using the formula n = m ÷ ___________, then divide each result by the reactant’s ___________ in the balanced equation. The reactant with the smaller quotient is limiting. The maximum amount of product that can form, calculated from the limiting reagent only, is the ___________.
4. Short-answer recall
Answer each question in 1–2 sentences using precise lesson terms. 8 marks (2 each)
4.1 Why is it incorrect to identify the limiting reagent simply by comparing the masses or raw moles of the two reactants?
4.2 In the reaction 2H2 + O2 → 2H2O, if you have 3 mol H2 and 2 mol O2, what quotient do you calculate for each reactant and which is limiting?
4.3 Once you have identified the limiting reagent, what is the next step in calculating the theoretical yield of a product?
4.4 Write the formula used to calculate the mass of excess reagent remaining after a reaction is complete.
5. Put the steps in order
The five steps for a full limiting-reagent calculation are listed below in the wrong order. Number them 1–5 in the boxes on the left to show the correct sequence. 4 marks
| Order | Step |
|---|---|
| Calculate the theoretical yield: use the limiting reagent’s moles and the mole ratio to product; then m = n × MM. | |
| Convert the mass of each reactant to moles using n = m ÷ MM. | |
| Confirm the equation is balanced and note the coefficient of each reactant and product. | |
| Identify the limiting reagent: the reactant with the smaller n ÷ coefficient quotient. | |
| Divide each reactant’s moles by its stoichiometric coefficient and compare the results. |
Q1 — Term–definition match
1.1 limiting reagent • 1.2 excess reagent • 1.3 theoretical yield • 1.4 stoichiometric ratio • 1.5 mole • 1.6 molar mass • 1.7 n ÷ coefficient comparison • 1.8 excess remaining.
Q2 — True / false with correction
2.1 False. The limiting reagent is identified by comparing n ÷ coefficient for each reactant — not by mass alone. A larger mass may belong to a heavy reactant that still provides more reaction-units than a lighter reactant.
2.2 True.
2.3 False. Theoretical yield must be calculated from the limiting reagent only. Using the excess reagent gives a larger, incorrect answer because the excess reagent has more reaction-units than the limiting reagent can actually consume.
2.4 False. The excess reagent is the one that remains when the reaction stops. It is the limiting reagent that runs out first.
2.5 True.
2.6 False. Raw moles alone cannot be compared directly without dividing by each coefficient first. For 2Na + Cl2 → 2NaCl: Na quotient = 0.44 ÷ 2 = 0.220; Cl2 quotient = 0.28 ÷ 1 = 0.280. Na has the smaller quotient, so Na is the limiting reagent — not Cl2 as a raw mole comparison might suggest.
Q3 — Cloze paragraph
In order: stoichiometric / consumed / limiting / excess / moles / molar mass / coefficient / theoretical yield.
Q4.1 — Why raw mass/moles comparison fails
Different substances have different molar masses and different stoichiometric coefficients. A reactant with a larger mass might have fewer moles, and a reactant with more moles might still not be limiting if its coefficient is larger. The only valid comparison is n ÷ coefficient for each reactant.
Q4.2 — H2 + O2 quotients
H2: 3 mol ÷ 2 = 1.50 • O2: 2 mol ÷ 1 = 2.00. H2 has the smaller quotient (1.50 < 2.00), so H2 is the limiting reagent.
Q4.3 — Next step after identifying LR
Use the moles of the limiting reagent and the mole ratio between the limiting reagent and the product (from the balanced equation coefficients) to calculate the moles of product: n(product) = n(LR) × [coeff(product) ÷ coeff(LR)]. Then convert to mass: m = n × MM(product).
Q4.4 — Excess remaining formula
n(excess remaining) = n(excess original) − n(excess consumed), where n(excess consumed) = n(LR) × [coeff(excess) ÷ coeff(LR)]. Then m(excess remaining) = n(excess remaining) × MM(excess).
Q5 — Correct step order
1 → Confirm balanced equation and note coefficients. 2 → Convert masses to moles (n = m ÷ MM). 3 → Divide each reactant’s moles by its coefficient. 4 → Identify LR (smaller quotient). 5 → Calculate theoretical yield from LR moles and mole ratio.
Marking criteria (4 marks): 1 mark for steps 1–2 in correct relative order; 1 mark for steps 2–3 in correct relative order; 1 mark for steps 3–4 in correct relative order; 1 mark for step 5 correctly placed last. Award 4 marks only if the full sequence is: Balance → n = m/MM → n ÷ coeff → identify LR → calculate yield.