Use the mole concept and balanced equations to perform stoichiometric calculations, including limiting reagent and percentage yield.

The mole, molar mass, the Avogadro constant, mole-ratio calculations from balanced equations, limiting reagent, and percentage yield.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-30

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What this dot point is asking

You must move confidently between mass, moles and particles and apply mole ratios from balanced equations.

The mole and molar mass

The central conversion is $n = m / M$ , where $n$ is moles, $m$ is mass in grams and $M$ is molar mass. The number of particles is $N = n \times 6.02 \times 10^{23}$ .

Mole-ratio calculations

A balanced equation gives the ratio in which substances react. The routine is:

Convert the known mass or volume to moles.
Multiply by the mole ratio from the balanced equation.
Convert the resulting moles back to the quantity asked for.

Limiting reagent and yield

When two amounts are given, one usually runs out first. Find moles of each reactant, divide by its coefficient, and the smallest value identifies the limiting reagent, which controls the amount of product.

The theoretical yield is the maximum product calculated from the limiting reagent. The percentage yield compares the actual product obtained:

In the exam, convert to moles first, identify the limiting reagent when two amounts are given, apply the mole ratio, and use percentage yield to compare your theoretical answer with any actual yield supplied.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 TASC5 marksA manufacturer produced 58.0 kg of ammonia by reacting 63.0 kg of nitrogen with 15.0 kg of hydrogen: N2(g) + 3H2(g) -> 2NH3(g). (M(H2) = 2.016, M(N2) = 28.02, M(NH3) = 17.034 g mol-1.) Show that the maximum possible yield of ammonia, if the reaction went to completion, is approximately 77.0 kg, then determine the percentage yield.

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Find the limiting reagent. n(N2) = 63 000 / 28.02 = 2248 mol; n(H2) = 15 000 / 2.016 = 7440 mol. The ratio needed is 1 N2 : 3 H2. For 2248 mol N2 you would need 6744 mol H2, and 7440 mol is available, so N2 is limiting. (Check H2: 7440/3 = 2480 mol N2 needed, more than available.) (2 marks)

Maximum ammonia. n(NH3) = 2 x n(N2) = 2 x 2248 = 4496 mol. Mass = 4496 x 17.034 = 76 600 g = 76.6 kg, approximately 77.0 kg. (2 marks)

Percentage yield = (actual / theoretical) x 100 = (58.0 / 76.6) x 100 = 75.7%. (1 mark)

2022 TASC3 marksA geochemist analysed a 100.0 g sample of hydrated iron carbonate and found it contained 24.94% Fe, 5.36% C, 5.40% H and 64.30% O. Calculate the empirical formula, including the degree of hydration, of the hydrated iron carbonate.

Show worked answer →

Treat each percentage as grams in 100.0 g and divide by molar mass to get moles. Fe: 24.94/55.85 = 0.4466; C: 5.36/12.01 = 0.4463; H: 5.40/1.008 = 5.357; O: 64.30/16.00 = 4.019.

Divide each by the smallest (0.4463): Fe = 1.00; C = 1.00; H = 12.0; O = 9.0.

So the formula is FeCH12O9. Rewriting as a carbonate plus water: the CO3 uses 3 O, leaving 6 O and 12 H, which is 6 H2O. So the empirical formula is FeCO3.6H2O (iron carbonate hexahydrate). (3 marks: moles, ratio, identifying the degree of hydration.)

2021 TASC3 marksOxygen gas is produced when potassium chlorate is heated: 2KClO3(s) -> 2KCl(s) + 3O2(g). 50.2 mL of gas is collected at 20.0 degrees C and 98.2 kPa. Calculate the mass of potassium chlorate used, assuming the reaction goes to completion. (Mr(KClO3) = 122.5.)

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Find moles of O2 using PV = nRT. P = 98.2 kPa = 98 200 Pa; V = 50.2 mL = 5.02 x 10-5 m3; T = 20.0 + 273 = 293 K; R = 8.314.

n(O2) = PV / RT = (98 200 x 5.02 x 10-5) / (8.314 x 293) = 4.930 / 2436 = 2.024 x 10-3 mol. (1 mark)

Mole ratio: 2 KClO3 produce 3 O2, so n(KClO3) = (2/3) x n(O2) = (2/3) x 2.024 x 10-3 = 1.349 x 10-3 mol. (1 mark)

Mass = n x Mr = 1.349 x 10-3 x 122.5 = 0.165 g. (1 mark)

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