ChemistryICSE Class 10 · Mole Concept and Stoichiometry

Calculations from Chemical Equations: Mass, Volume and Gay-Lussac's Law (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

A balanced equation gives the ratio in which substances react, in moles. To calculate, convert the given quantity to moles, use the mole ratio from the equation, then convert back to grams (moles × molar mass) or to litres at S.T.P. (moles × 22.4 L). For gases at the same temperature and pressure, volumes follow the coefficients directly.

Every stoichiometry question crosses the same bridge: convert to moles, apply the equation's mole ratio, then convert back to mass or volume.

Illustration drawn for QuestionBuddy.

Understand Calculations from Chemical Equations

A balanced chemical equation is a recipe written in moles. In $C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2}$ , the coefficients say that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid and gives 1 mole of carbon dioxide. Because 1 mole has a fixed mass (the gram molecular mass) and, for a gas, a fixed volume at S.T.P. (22.4 L), every mass or volume question becomes a mole question.

Three kinds of calculation.

1. Mass to mass. Convert the given mass to moles ( $n = m / M$ ), multiply by the mole ratio from the equation, then convert to mass of the required substance.
2. Mass to volume (or volume to mass). Same route, but use 22.4 L per mole for the gas at S.T.P. (0 °C, 1 atm).
3. Volume to volume (Gay-Lussac's law). When gases react, the volumes of the gaseous reactants and products, measured at the same temperature and pressure, are in a simple whole-number ratio, the ratio of their coefficients. So in $N_{2} + 3 H_{2} \to 2 N H_{3}$ , 10 cm³ of nitrogen needs 30 cm³ of hydrogen and gives 20 cm³ of ammonia. No molar masses are needed. Gay-Lussac's law applies only to gases: water formed in a combustion is a liquid once the mixture is cooled to room temperature, so it is left out of the gas volume.

Gas left over. If one gas is in excess, only the amount the equation allows reacts. The final gas mixture is the unused excess plus the gaseous products. If the mixture is then passed through caustic potash (KOH) solution, carbon dioxide is absorbed and its volume drops out.

Which reactant runs out? When masses of two reactants are given, find the moles of each and compare with the ratio in the equation. The one that is used up first (the limiting reactant) decides how much product forms; the other is left in excess.

Why the method works. Avogadro's law (equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules) links the coefficients, which count molecules, to volumes of gases. Molar mass links them to grams. The syllabus asks for simple calculations of this kind on the weight and/or volume of reactants and products.

Calculations from Chemical Equations: formulas and key facts

Moles from mass: $n = \frac{m}{M}$; $m$ in g, $M$ = gram molecular (or atomic) mass in g mol⁻¹.
Moles from gas volume at S.T.P.: $n = \frac{V}{22.4 L}$; 22.4 L = 22 400 cm³; only for gases at S.T.P.
Mole ratio from the equation: $\frac{n _{wanted}}{n _{given}} = \frac{coefficient of wanted}{coefficient of given}$; Always from the balanced equation.
Gay-Lussac's law (gases, same T and P): $V_{1} : V_{2} : V_{3} = a : b : c$; $a$ , $b$ , $c$ are the coefficients of the gaseous substances.
The route for every numerical: 1.Given quantity
2.moles of given
3.mole ratio
4.moles of wanted
5.grams or litres
Useful molar masses (g mol⁻¹): Calcium carbonate CaCO₃ 100
Magnesium carbonate MgCO₃ 84
Sodium hydrogen carbonate NaHCO₃ 84
Sodium carbonate Na₂CO₃ 106
Hydrogen chloride HCl 36.5
Sulphuric acid H₂SO₄ 98
Potassium nitrate KNO₃ 101
Calcium hydroxide Ca(OH)₂ 74

Calcium carbonate CaCO₃	100
Magnesium carbonate MgCO₃	84
Sodium hydrogen carbonate NaHCO₃	84
Sodium carbonate Na₂CO₃	106
Hydrogen chloride HCl	36.5
Sulphuric acid H₂SO₄	98
Potassium nitrate KNO₃	101
Calcium hydroxide Ca(OH)₂	74

How to solve Calculations from Chemical Equations questions

Every stoichiometry numerical follows the same four lines. Write them out; the examiner gives marks for each line, not only the answer.

1Write the balanced equation (or check the one given) and underline the two substances the question links.
2Convert the given quantity to moles: mass ÷ molar mass, or volume at S.T.P. ÷ 22.4 L.
3Use the coefficients to get moles of the wanted substance.
4Convert to what is asked: moles × molar mass for grams, moles × 22.4 L for a gas at S.T.P., moles × 6.022 × 10²³ for particles.
5For gas volumes at the same temperature and pressure, skip moles: multiply the given volume by the ratio of coefficients.
6If two reactant amounts are given, check which runs out first before calculating the product.
7Finish with the unit and, for gases, 'at S.T.P.'

Calculations from Chemical Equations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Mass to volume: carbon dioxide from a fuel

Suppose a camping stove canister holds 116 g of butane, C₄H₁₀, which burns completely: $2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$ . (C = 12, H = 1; take volumes at S.T.P.)

What volume of oxygen does the stove use, and what volume of carbon dioxide does it release, when the whole canister is burnt?

Try it first, then show the solutionSolution

Molar mass of butane $= 4 \times 12 + 10 \times 1 = 58$ g mol⁻¹; moles $= 116/58 = 2$ mol.
From the equation, 2 mol butane needs 13 mol O₂ and gives 8 mol CO₂.
O₂: $13 \times 22.4 = 291.2$ L. CO₂: $8 \times 22.4 = 179.2$ L.

Answer: About 291 L of oxygen used and 179 L of carbon dioxide released at S.T.P., which is why stoves must never be used in a closed tent.

Memory hook: Grams in, moles through the middle, litres out.

Mass to mass: how much acid an antacid neutralises

Suppose an antacid tablet contains 0.29 g of magnesium hydroxide, which neutralises stomach acid: $M g (O H)_{2} + 2 H C l \to M g C l_{2} + 2 H_{2} O$ . (Mg = 24, O = 16, H = 1, Cl = 35.5)

What mass of hydrochloric acid (HCl) can one tablet neutralise?

Try it first, then show the solutionSolution

Molar mass of Mg(OH)₂ $= 24 + 2 \times 17 = 58$ ; moles $= 0.29/58 = 0.005$ mol.
Mole ratio Mg(OH)₂ : HCl = 1 : 2, so moles of HCl $= 0.010$ mol.
Mass of HCl $= 0.010 \times 36.5 = 0.365$ g.

Answer: 0.365 g of HCl per tablet.

Memory hook: The coefficients are the exchange rate between substances.

Tricks to remember Calculations from Chemical Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

GMR-G: Given, Moles, Ratio, Get
Given → Moles → Ratio → Get. Four steps, four lines of working.
Use it when: Any mass or volume numerical.
Gases only, no grams
If all the quantities are gas volumes at the same T and P, use the coefficients as volume ratios. No molar mass, no 22.4.
Use it when: Gay-Lussac's law questions with cm³ data.
Cool it, drop the water
After combustion 'cooled to room temperature' means steam has condensed: water is not counted in the gas volume.
Use it when: Total volume of gases left after burning a hydrocarbon.

Calculations from Chemical Equations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Ammonia from hydrogen · 1 mark · MCQBoard level

In the reaction $N_{2} + 3 H_{2} \to 2 N H_{3}$ , 60 cm³ of hydrogen reacts completely with excess nitrogen. The volume of ammonia formed at the same temperature and pressure is:
(a) 20 cm³
(b) 40 cm³
(c) 60 cm³
(d) 120 cm³

Try it first, then show the solutionSolution

Volume ratio H₂ : NH₃ = 3 : 2 (Gay-Lussac's law).
Volume of NH₃ $= 60 \times \frac{2}{3} = 40$ cm³.

Answer: (b) 40 cm³.

Example 2: Heating magnesium carbonate · 2 marks · Short answerBoard level

Magnesium carbonate decomposes on heating: $M g C O_{3} Δ M g O + C O_{2}$ . Calculate the volume of carbon dioxide at S.T.P. and the mass of magnesium oxide obtained from 21 g of magnesium carbonate. (Mg = 24, C = 12, O = 16)

Try it first, then show the solutionSolution

Molar mass of MgCO₃ $= 24 + 12 + 48 = 84$ ; moles $= 21/84 = 0.25$ mol.
Mole ratio 1 : 1 : 1, so 0.25 mol CO₂ and 0.25 mol MgO.
CO₂: $0.25 \times 22.4 = 5.6$ L at S.T.P. MgO: $0.25 \times 40 = 10$ g.

Answer: 5.6 L of CO₂ at S.T.P. and 10 g of MgO.

Example 3: Aluminium and hydrochloric acid · 3 marks · Short answerHarder board level

Aluminium reacts with dilute hydrochloric acid: $2 A l + 6 H C l \to 2 A l C l_{3} + 3 H_{2}$ . For 8.1 g of aluminium, calculate (a) the volume of hydrogen at S.T.P., (b) the mass of HCl used, (c) the mass of aluminium chloride formed. (Al = 27, H = 1, Cl = 35.5)

Try it first, then show the solutionSolution

Moles of Al $= 8.1/27 = 0.3$ mol.
(a) Al : H₂ = 2 : 3, so H₂ $= 0.3 \times \frac{3}{2} = 0.45$ mol $= 0.45 \times 22.4 = 10.08$ L.
(b) Al : HCl = 2 : 6, so HCl $= 0.9$ mol $= 0.9 \times 36.5 = 32.85$ g.
(c) Al : AlCl₃ = 1 : 1, so 0.3 mol $\times 133.5 = 40.05$ g.

Answer: (a) 10.08 L (b) 32.85 g (c) 40.05 g.

Example 4: Burning propane in excess oxygen · 3 marks · Short answerHarder board level

20 cm³ of propane is burnt in 150 cm³ of oxygen: $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ . All volumes are measured at room temperature and pressure. Find (a) the volume of oxygen left unused, (b) the total volume of gases after the reaction, (c) the volume remaining after the gases are passed through caustic potash solution.

Try it first, then show the solutionSolution

Oxygen needed $= 20 \times 5 = 100$ cm³; oxygen left $= 150 - 100 = 50$ cm³.
CO₂ formed $= 20 \times 3 = 60$ cm³. Water is a liquid at room temperature, so it is not counted.
Total gas $= 50 + 60 = 110$ cm³.
KOH absorbs CO₂, leaving only the 50 cm³ of oxygen.

Answer: (a) 50 cm³ O₂ (b) 110 cm³ (c) 50 cm³ (oxygen).

Example 5: Which reactant runs out? · 4 marks · Long answerHOTS / competency

4.8 g of magnesium is added to a solution containing 10.95 g of hydrogen chloride: $M g + 2 H C l \to M g C l_{2} + H_{2}$ . (a) Which reactant is left over, and what mass of it remains? (b) What volume of hydrogen forms at S.T.P.? (Mg = 24, H = 1, Cl = 35.5)

Try it first, then show the solutionSolution

Moles of Mg $= 4.8/24 = 0.2$ ; moles of HCl $= 10.95/36.5 = 0.3$ .
0.2 mol Mg would need 0.4 mol HCl, but only 0.3 mol is present: HCl runs out first (limiting).
Mg used $= 0.3/2 = 0.15$ mol $= 3.6$ g; Mg left $= 4.8 - 3.6 = 1.2$ g.
H₂ formed $= 0.15$ mol $= 0.15 \times 22.4 = 3.36$ L at S.T.P.

Answer: (a) Magnesium is in excess; 1.2 g remains. (b) 3.36 L of hydrogen at S.T.P.

Example 6: Formula of a gas from volumes · 3 marks · Short answerHarder board level

10 cm³ of a gaseous hydrocarbon CₓHᵧ needs exactly 45 cm³ of oxygen for complete combustion and gives 30 cm³ of carbon dioxide, all at the same temperature and pressure. Use Gay-Lussac's law to find its molecular formula.

Try it first, then show the solutionSolution

General equation: $C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$ .
CO₂ : hydrocarbon $= 30 : 10$ , so $x = 3$ .
O₂ : hydrocarbon $= 45 : 10 = 4.5$ , so $x + \frac{y}{4} = 4.5$ , giving $y = 6$ .
Check: $2 C_{3} H_{6} + 9 O_{2} \to 6 C O_{2} + 6 H_{2} O$ ; 10 cm³ needs 45 cm³ O₂.

Answer: C₃H₆ (propene).

Where students lose marks in Calculations from Chemical Equations

The slips examiners see most often, and what to write instead.

Mistake: Writing only the final number
In 2024 (Q5(iii)) and 2025 (Q3(iii)) examiners reported answers such as 500 g or 98 g with no working, or a wrong molar mass substituted.
Wrong working
Mass of 5 moles of CaCO₃ = 500 g
Right working
Mass = moles × molar mass = 5 × 100 = 500 g
Do this: Write the moles, the ratio and the conversion as separate lines with units.
Usually costs: Method marks, often 1 to 2.
Mistake: Forgetting to multiply by the number of moles
Examiners noted 22.4 L written instead of 5 × 22.4 L for carbon dioxide (2024 Q5(iii)).
Do this: 22.4 L is the volume of one mole only. Multiply by the moles you found.
Usually costs: 1 mark.
Mistake: Total volume of gases: counting only the products
In 2025 (Q1(xv)) many candidates gave the volume of gases formed, not the total left, which includes the unused excess reactant gas.
Wrong working
Gas left = volume of CO₂ only
Right working
Gas left = unused excess gas + gaseous products (water excluded when cooled)
Do this: List each gas after the reaction: excess reactant, gaseous products. Leave out liquid water.
Usually costs: 1 mark (MCQ lost entirely).
Mistake: Reading the mole ratio wrongly
Examiners reported errors reading the HCl ratio from $C a C O_{3} + 2 H C l$ (2024 Q5(iii)).
Do this: Copy the coefficients of the two substances you are linking above your working before calculating.
Usually costs: 1 mark and every mark that depends on it.

Exam tips

Use 22.4 L (22 400 cm³) per mole only for gases at S.T.P.; for volume-to-volume data at the same temperature and pressure, use the coefficient ratio instead.
Use the atomic masses printed in the question, even if you remember slightly different values.
When two reactant quantities are given, test which one runs out before calculating any product.
End every gas answer with 'at S.T.P.' when the question is about S.T.P.
Keep three significant figures in intermediate moles to avoid rounding drift.

Practice questions on Calculations from Chemical Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The number of moles of oxygen needed to burn 2 moles of methane completely ( $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ ) is:
(a) 1
(b) 2
(c) 4
(d) 8
Hint
Read the coefficient ratio.
Answer
(c) 4.
Q2 · 1 mark · Very short answerBoard level
State Gay-Lussac's law of combining volumes.
Hint
Gases, same temperature and pressure.
Answer
When gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the volumes of the gaseous products, all measured at the same temperature and pressure.
Q3 · 2 marks · Short answerHarder board level
Potassium nitrate decomposes on heating: $2 K N O_{3} Δ 2 K N O_{2} + O_{2}$ . What mass of potassium nitrate must be heated to give 1.12 L of oxygen at S.T.P.? (K = 39, N = 14, O = 16)
Hint
Moles of O₂ first, then double.
Answer
Moles of O₂ $= 1.12/22.4 = 0.05$ ; moles of KNO₃ $= 0.10$ ; mass $= 0.10 \times 101 = 10.1$ g.
Q4 · 3 marks · Short answerHarder board level
Calcium reacts with water: $C a + 2 H_{2} O \to C a (O H)_{2} + H_{2}$ . For 2 g of calcium, find (a) the volume of hydrogen at S.T.P. and (b) the mass of calcium hydroxide formed. (Ca = 40, O = 16, H = 1)
Hint
Ca : H₂ : Ca(OH)₂ = 1 : 1 : 1.
Answer
Moles of Ca $= 2/40 = 0.05$ . (a) H₂ $= 0.05 \times 22.4 = 1.12$ L. (b) Ca(OH)₂ $= 0.05 \times 74 = 3.7$ g.
Q5 · 3 marks · Short answerHarder board level
40 cm³ of hydrogen sulphide is burnt in 80 cm³ of oxygen: $2 H_{2} S + 3 O_{2} \to 2 H_{2} O + 2 S O_{2}$ . The products are cooled to room temperature. Find the volume and composition of the gas mixture left.
Hint
Water condenses; count only gases.
Answer
O₂ used $= 40 \times \frac{3}{2} = 60$ cm³, so 20 cm³ O₂ is left. SO₂ formed $= 40$ cm³. Total $= 60$ cm³ (20 cm³ O₂ + 40 cm³ SO₂).
Q6 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In $N_{2} + 3 H_{2} \to 2 N H_{3}$ , 1 litre of nitrogen gives 2 litres of ammonia at the same temperature and pressure.
Reason (R): Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Avogadro's law turns molecule ratios into volume ratios.
Answer
(a). The equation gives 2 molecules of NH₃ per molecule of N₂; by Avogadro's law that means 2 volumes per volume.
Q7 · 3 marks · Short answerHOTS / competency
A student heats 8.4 g of sodium hydrogen carbonate until no more gas is given off: $2 N a H C O_{3} Δ N a_{2} C O_{3} + H_{2} O + C O_{2}$ . (a) What mass of residue should she expect? (b) By how much should the mass fall? (c) What volume of CO₂ is released at S.T.P.? (Na = 23, H = 1, C = 12, O = 16)
Hint
2 mol NaHCO₃ (168 g) gives 1 mol Na₂CO₃ (106 g).
Answer
Moles of NaHCO₃ $= 8.4/84 = 0.1$ . (a) Na₂CO₃ $= 0.05$ mol $= 5.3$ g. (b) Loss $= 8.4 - 5.3 = 3.1$ g (water and CO₂ escape). (c) CO₂ $= 0.05 \times 22.4 = 1.12$ L.
Q8 · 3 marks · Short answerHOTS / competency
A 12.5 g sample of impure limestone reacts completely with excess dilute hydrochloric acid and gives 2.24 L of carbon dioxide at S.T.P.: $C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2}$ . Assuming the impurities give no gas, find the percentage of calcium carbonate in the sample. (Ca = 40, C = 12, O = 16)
Hint
Work back from the gas to the mass of pure CaCO₃.
Answer
Moles of CO₂ $= 2.24/22.4 = 0.1$ = moles of CaCO₃; mass $= 0.1 \times 100 = 10$ g. Percentage $= \frac{10}{12.5} \times 100 = 80%$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Volume of 0.5 mol of any gas at S.T.P.?Show answer
11.2 L.
2.In $2 H_{2} + O_{2} \to 2 H_{2} O$ , what volume of oxygen reacts with 50 cm³ of hydrogen?Show answer
25 cm³ (ratio 2 : 1).
3.Why is water ignored in the final gas volume after cooling?Show answer
At room temperature it is a liquid; Gay-Lussac's law applies only to gases.
4.Which solution absorbs carbon dioxide from a gas mixture?Show answer
Caustic potash (KOH) solution.
5.Moles in 9.8 g of H₂SO₄ (M = 98)?Show answer
0.1 mol.
6.What is a limiting reactant?Show answer
The reactant used up first, which fixes the amount of product.

Calculations from Chemical Equations: frequently asked questions

How do you calculate the volume of a gas from a mass in ICSE chemistry?

Convert the mass to moles (mass ÷ molar mass), use the mole ratio from the balanced equation to get moles of the gas, then multiply by 22.4 L for the volume at S.T.P.

When can I use Gay-Lussac's law instead of moles?

When all the quantities in the question are volumes of gases measured at the same temperature and pressure. Then volumes are in the same ratio as the coefficients.

Do I count steam in the volume of gases after a combustion?

Only if the question says the volume is measured above 100 °C. If the gases are cooled to room temperature, water is a liquid and is not counted.

What is the most common mistake in stoichiometry numericals?

Skipping the working. ICSE examiners repeatedly note bare answers and wrong mole ratios; showing moles, ratio and conversion earns method marks even if the arithmetic slips.