A-Level Chemistry 9701 / Topic 2 / AS

The equation counts particles.

A balanced equation gives a ratio of moles, not of grams. Convert what you know into moles, use the ratio, convert back. When two reactants are given, one runs out first and decides how much product there can be.

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The key idea

Mass of A → moles of A (divide by M) → moles of B (use the equation's ratio) → mass of B (multiply by M). The limiting reagent is the reactant that is completely used up; it sets the maximum amount of product, and the other reactant is in excess. Percentage yield = actual yield ÷ theoretical yield × 100, with the theoretical yield calculated from the limiting reagent.

Fig. 1: the three-step route through the mole

Section 01

The three-step route.

Never compare grams directly: 1 g of one substance and 1 g of another contain different numbers of particles. Go through moles.

Question: what mass of calcium oxide forms when 25.0 g of calcium carbonate is heated? CaCO₃ → CaO + CO₂.
Step 1: n(CaCO₃) = 25.0 ÷ 100.1 = 0.250 mol.
Step 2: the ratio is 1 : 1, so n(CaO) = 0.250 mol.
Step 3: mass = 0.250 × 56.1 = 14.0 g. The other 11.0 g left as CO₂ gas.

Stage 1 · Learn

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Four quick checks. Each correct answer earns XP and records this skill on your profile.

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In 2Mg + O₂ → 2MgO, how many moles of MgO form from 0.30 mol of Mg (with excess oxygen)?

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In N₂ + 3H₂ → 2NH₃, how many moles of H₂ react with 0.50 mol of N₂?

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The limiting reagent is the reactant that:

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The theoretical yield of a product is 10.0 g and 8.0 g is obtained. The percentage yield is:

Section 02

Limiting and excess.

Set the two amounts in the simulator and watch which bar empties first. For each reactant, divide the moles you have by its coefficient in the equation: the smaller answer is the limiting reagent.

Reaction: Mg + 2HCl → MgCl₂ + H₂ with 0.10 mol Mg and 0.15 mol HCl.
Test: Mg 0.10 ÷ 1 = 0.10; HCl 0.15 ÷ 2 = 0.075. HCl gives the smaller number, so HCl is limiting.
Product: n(H₂) = 0.15 ÷ 2 = 0.075 mol.
Excess: Mg used = 0.075 mol, so 0.10 − 0.075 = 0.025 mol of magnesium is left over.

Section 03

Percentage yield.

Change the mass of reactant and the mass of product actually collected, and read the percentage yield. Then try to explain a yield below 100% in words.

Theoretical yield: the mass of product calculated from the limiting reagent, assuming complete reaction.
Example: 25.0 g of CaCO₃ should give 14.0 g of CaO. If 11.2 g is collected, the yield is 11.2 ÷ 14.0 × 100 = 80.0%.
Why below 100%: the reaction is incomplete or reversible, side reactions form other products, and product is lost when it is separated and purified.

Examiner trap

Use the ratio from the equation, not a ratio of masses, and always work the theoretical yield from the limiting reagent. If you use the reagent in excess, the answer will be too big and the percentage yield too small.

Stage 2 · Exam

Exam-style questions

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Iron(III) oxide is reduced by excess carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. What mass of iron is produced from 80.0 g of Fe₂O₃? (M_r: Fe₂O₃ 159.6; A_r: Fe 55.8)

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0.200 mol of magnesium is added to a solution containing 0.300 mol of HCl. Mg + 2HCl → MgCl₂ + H₂. Which statement is correct?

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Titanium is extracted by TiCl₄ + 2Mg → Ti + 2MgCl₂. What mass of magnesium is needed to produce 1.00 kg of titanium? (A_r: Mg 24.3, Ti 47.9)

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10.0 g of ethanol (M_r 46.0) is oxidised to ethanoic acid (M_r 60.0), one mole of acid per mole of ethanol. 12.0 g of ethanoic acid is obtained. What is the percentage yield?

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Six original Cambridge-style questions

Reacting masses from equations, finding the limiting and excess reagent, percentage yield, deducing an equation from masses, and an industrial-scale calculation.