TASC TCE 2023-style practice: A manufacturer produced 58.0\ kg of ammonia by reacting 63.0\ kg of nitrogen with 15.0\ kg of hydrogen: N_2(g) + 3H_2(g) → 2NH_3(g). (M(H_2) = 2.016, M(N_2) = 28.02, M(NH_3) = 17.03\ 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.

Find the limiting reagent. n(N_2) = 63\,000 / 28.02 = 2248\ mol; n(H_2) = 15\,000 / 2.016 = 7440\ mol. The ratio needed is 1\ N_2 : 3\ H_2. For 2248\ mol of N_2 you would need 6744\ mol of H_2, and 7440\ mol is available, so N_2 is limiting. (2 marks) Maximum ammonia: n(NH_3) = 2 × n(N_2) = 2 × 2248 = 4496\ mol. Mass = 4496 × 17.03 = 76\,600\ g = 76.6\ kg, approximately 77.0\ kg. (2 marks) Percentage yield = (actual / theoretical) × 100 = (58.0 / 76.6) × 100 = 75.7\%. (1 mark)

TASC TCE 2022-style practice: A geochemist analysed a sample of a hydrated iron carbonate and found it contained 24.94\% Fe, 5.36\% C, 5.40\% H and 64.30\% O by mass. Calculate the empirical formula, including the degree of hydration. (M(Fe) = 55.85, M(C) = 12.01, M(H) = 1.008, M(O) = 16.00\ g mol^-1.)

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. (1 mark) Divide each by the smallest (0.4463): Fe = 1.00; C = 1.00; H = 12.0; O = 9.0. (1 mark) The ratio is FeCH_12O_9. Rewriting as a carbonate plus water, the CO_3 uses 3 of the O, leaving 6\ O and 12\ H, which is 6\ H_2O. So the empirical formula is FeCO_3 · 6H_2O (iron carbonate hexahydrate). (1 mark)

TASC TCE 2021-style practice: Oxygen gas is produced when potassium chlorate is heated: 2KClO_3(s) → 2KCl(s) + 3O_2(g). 50.2\ mL of gas is collected at 20.0\ ^C and 98.2\ kPa. Calculate the mass of potassium chlorate used, assuming the reaction goes to completion. (M(KClO_3) = 122.5\ g mol^-1, R = 8.314\ J K^-1mol^-1.)

Find moles of O_2 using pV = nRT. p = 98.2\ kPa = 98\,200\ Pa; V = 50.2\ mL = 5.02 × 10^-5\ m^3; T = 20.0 + 273 = 293\ K. $n(O_2) = (pV)/(RT) = 98\,200 × 5.02 × 10^-58.314 × 293 = 2.02 × 10^-3\ mol.$ (1 mark) Mole ratio: 2\ KClO_3 produce 3\ O_2, so n(KClO_3) = 23 × n(O_2) = 23 × 2.02 × 10^-3 = 1.35 × 10^-3\ mol. (1 mark) Mass = n × M = 1.35 × 10^-3 × 122.5 = 0.165\ g. (1 mark)

§-Syllabus dot point

TASChemistrySyllabus dot point

How do we relate amounts of reactants and products in a reaction?

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

The mole, molar mass and the Avogadro constant, mole-ratio calculations from balanced equations, limiting reagent, empirical formulae and percentage yield, with fully worked TASC-style numerical examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

TASC expects you to move confidently between mass, moles, number of particles and gas quantities, and to apply mole ratios drawn straight from a balanced equation. Almost every quantitative question in Unit 3 begins with a mole calculation, so this is the foundation skill for titrations, gas laws, thermochemistry and equilibrium.

The mole and molar mass

The three conversions you must know by heart are:

mass to moles: $n = \dfrac{m}{M}$
particles to moles: $n = \dfrac{N}{N_A}$ , so $N = n \times 6.022 \times 10^{23}$
moles of gas to volume: at STP a gas occupies $V_m = 22.71\ \text{L mol}^{-1}$ , so $n = \dfrac{V}{V_m}$ (TASC uses $0\ ^\circ\text{C}$ , $100\ \text{kPa}$ for STP; check your data sheet).

Mole-ratio calculations

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

Write a balanced equation.
Convert the known quantity (mass, volume or concentration) to moles.
Multiply by the mole ratio (coefficient of wanted species over coefficient of known species).
Convert the resulting moles back to the quantity asked for.

Mass of product from a balanced equation

Calculate the mass of magnesium oxide that forms when $6.0\ \text{g}$ of magnesium burns completely in oxygen. The balanced equation is $2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$ , with $M(\text{Mg}) = 24.31\ \text{g mol}^{-1}$ and $M(\text{MgO}) = 40.31\ \text{g mol}^{-1}$ .

Step 1: Convert the known mass of magnesium to moles

Working in moles is essential because the balanced equation gives a ratio of moles, not grams. Divide the mass by the molar mass of magnesium:

n(\text{Mg}) = \frac{m}{M} = \frac{6.0}{24.31} = 0.247\ \text{mol}.

Step 2: Apply the stoichiometric mole ratio

Read the ratio directly from the coefficients in the balanced equation. The equation shows $2\ \text{Mg}$ producing $2\ \text{MgO}$ , a $1:1$ ratio. Therefore the moles of magnesium oxide equal the moles of magnesium:

n(\text{MgO}) = 0.247\ \text{mol}.

Step 3: Convert moles of product back to mass

Multiply the moles of magnesium oxide by its molar mass to obtain the final answer:

m(\text{MgO}) = nM = 0.247 \times 40.31 = 9.95\ \text{g}.

As a sense check, the oxygen taken up is $0.247 \times 16.00 = 3.95\ \text{g}$ , and $6.0 + 3.95 = 9.95\ \text{g}$ , confirming mass is conserved.

Final answer: 9.95 g of magnesium oxide forms.

Limiting reagent

When two reactant amounts are given, one usually runs out first. Find moles of each reactant, divide each by its coefficient, and the smallest value identifies the limiting reagent, which controls how much product forms. The other reactant is in excess and some is left over.

Percentage yield and atom economy

The theoretical yield is the maximum product calculated from the limiting reagent. The actual yield is always less because of side reactions, incomplete reaction, and losses in handling. Percentage yield compares the two:

Atom economy is a different idea used in green chemistry: it measures the fraction of reactant mass that ends up in the desired product, $\text{atom economy} = \dfrac{M(\text{desired product})}{M(\text{all products})} \times 100\%$ . A reaction can have high yield but poor atom economy if it makes large amounts of by-product.

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms. From percentage composition, treat each percentage as grams in $100\ \text{g}$ , divide by molar mass, then divide by the smallest result. The molecular formula is a whole-number multiple of the empirical formula, found from the relative molecular mass.

In the exam, convert to moles first, identify the limiting reagent whenever two amounts are given, apply the mole ratio from the balanced equation, and finish with the quantity asked for. Carry one or two extra significant figures through the working and round only at the end.

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.

TCE 20235 marksA manufacturer produced

58.0\ \text{kg}

of ammonia by reacting

63.0\ \text{kg}

of nitrogen with

15.0\ \text{kg}

of hydrogen:

\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)

. (

M(\text{H}_2) = 2.016

M(\text{N}_2) = 28.02

M(\text{NH}_3) = 17.03\ \text{g mol}^{-1}

.) Show that the maximum possible yield of ammonia, if the reaction went to completion, is approximately

77.0\ \text{kg}

, then determine the percentage yield.

Show worked answer →

Find the limiting reagent. $n(\text{N}_2) = 63\,000 / 28.02 = 2248\ \text{mol}$ ; $n(\text{H}_2) = 15\,000 / 2.016 = 7440\ \text{mol}$ . The ratio needed is $1\ \text{N}_2 : 3\ \text{H}_2$ . For $2248\ \text{mol}$ of $\text{N}_2$ you would need $6744\ \text{mol}$ of $\text{H}_2$ , and $7440\ \text{mol}$ is available, so $\text{N}_2$ is limiting. (2 marks)

Maximum ammonia: $n(\text{NH}_3) = 2 \times n(\text{N}_2) = 2 \times 2248 = 4496\ \text{mol}$ . Mass $= 4496 \times 17.03 = 76\,600\ \text{g} = 76.6\ \text{kg}$ , approximately $77.0\ \text{kg}$ . (2 marks)

Percentage yield $= (\text{actual} / \text{theoretical}) \times 100 = (58.0 / 76.6) \times 100 = 75.7\%$ . (1 mark)

TCE 20223 marksA geochemist analysed a sample of a hydrated iron carbonate and found it contained

24.94\%

\text{Fe}

5.36\%

\text{C}

5.40\%

\text{H}

and

64.30\%

\text{O}

by mass. Calculate the empirical formula, including the degree of hydration. (

M(\text{Fe}) = 55.85

M(\text{C}) = 12.01

M(\text{H}) = 1.008

M(\text{O}) = 16.00\ \text{g mol}^{-1}

Show worked answer →

Treat each percentage as grams in $100.0\ \text{g}$ and divide by molar mass to get moles. $\text{Fe}: 24.94/55.85 = 0.4466$ ; $\text{C}: 5.36/12.01 = 0.4463$ ; $\text{H}: 5.40/1.008 = 5.357$ ; $\text{O}: 64.30/16.00 = 4.019$ . (1 mark)

Divide each by the smallest ( $0.4463$ ): $\text{Fe} = 1.00$ ; $\text{C} = 1.00$ ; $\text{H} = 12.0$ ; $\text{O} = 9.0$ . (1 mark)

The ratio is $\text{FeCH}_{12}\text{O}_9$ . Rewriting as a carbonate plus water, the $\text{CO}_3$ uses $3$ of the $\text{O}$ , leaving $6\ \text{O}$ and $12\ \text{H}$ , which is $6\ \text{H}_2\text{O}$ . So the empirical formula is $\text{FeCO}_3 \cdot 6\text{H}_2\text{O}$ (iron carbonate hexahydrate). (1 mark)

TCE 20213 marksOxygen gas is produced when potassium chlorate is heated:

2\text{KClO}_3(s) \rightarrow 2\text{KCl}(s) + 3\text{O}_2(g)

50.2\ \text{mL}

of gas is collected at

20.0\ ^\circ\text{C}

and

98.2\ \text{kPa}

. Calculate the mass of potassium chlorate used, assuming the reaction goes to completion. (

M(\text{KClO}_3) = 122.5\ \text{g mol}^{-1}

R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}

Show worked answer →

Find moles of $\text{O}_2$ using $pV = nRT$ . $p = 98.2\ \text{kPa} = 98\,200\ \text{Pa}$ ; $V = 50.2\ \text{mL} = 5.02 \times 10^{-5}\ \text{m}^3$ ; $T = 20.0 + 273 = 293\ \text{K}$ .

n(\text{O}_2) = \frac{pV}{RT} = \frac{98\,200 \times 5.02 \times 10^{-5}}{8.314 \times 293} = 2.02 \times 10^{-3}\ \text{mol}.

(1 mark)

Mole ratio: $2\ \text{KClO}_3$ produce $3\ \text{O}_2$ , so $n(\text{KClO}_3) = \tfrac{2}{3} \times n(\text{O}_2) = \tfrac{2}{3} \times 2.02 \times 10^{-3} = 1.35 \times 10^{-3}\ \text{mol}$ . (1 mark)

Mass $= n \times M = 1.35 \times 10^{-3} \times 122.5 = 0.165\ \text{g}$ . (1 mark)