Chemistry

Solved stoichiometry exercises, from basic to exam level

Fourteen exercises with solutions: moles, composition, formulas, limiting reagent, yield and purity, acid-base and redox titrations, combustion analysis, mixtures, multi-stage processes, gases.

ielettronica 25 October 2026 7 min read 0 comments

Stoichiometry exercises of the kind set in university Chemistry courses, in three levels: basic, to fix the method; intermediate, with limiting reagent, yield, analysis and gases; exam, with the kind of problems found in written exams and mid-term tests. For each exercise the result is given straight away, so you can check yours; the full solution opens with a click. The method and the proofs are in the lesson and the in-depth study linked at the top.

Atomic masses used (g/mol): H 1.008; C 12.01; N 14.01; O 16.00; Na 22.99; Mg 24.31; Al 26.98; S 32.06; Cl 35.45; K 39.10; Ca 40.08; Fe 55.85; Mn 54.94; Ag 107.87. Gas constant $R = 0.082\,06$ L·atm/(mol·K).

Basic level

Exercise 1. Moles, formula units and atoms

How many moles, how many formula units and how many oxygen atoms are there in 25.0 g of calcium carbonate, CaCO₃?

Result: 0.250 mol; $1.50\cdot 10^{23}$ formula units; $4.51\cdot 10^{23}$ O atoms.

Solution

Molar mass: $M = 40.08 + 12.01 + 3\cdot 16.00 = 100.09$ g/mol.

Moles: $n = 25.0/100.09 = 0.2498$ mol, i.e. 0.250 mol to three significant figures.

Formula units: $N = nN_A = 0.2498\cdot 6.022\cdot 10^{23} = 1.504\cdot 10^{23}$ .

Each unit contains 3 oxygen atoms: $3\cdot 1.504\cdot 10^{23} = 4.51\cdot 10^{23}$ .

Exercise 2. Percentage composition

Compute the percentage composition by mass of ammonium sulfate, (NH₄)₂SO₄, a fertiliser.

Result: N 21.2 %; H 6.1 %; S 24.3 %; O 48.4 %.

Solution

The formula contains 2 N, 8 H, 1 S, 4 O. Molar mass: $2\cdot 14.01 + 8\cdot 1.008 + 32.06 + 4\cdot 16.00 = 28.02 + 8.064 + 32.06 + 64.00 = 132.14$ g/mol.

Percentages: N $28.02/132.14 = 21.2$ %; H $8.064/132.14 = 6.1$ %; S $32.06/132.14 = 24.3$ %; O $64.00/132.14 = 48.4$ %. The sum is 100.0 %: the check to make every time.

Exercise 3. Combustion of propane

22.0 g of propane, C₃H₈, are burned. How many grams of oxygen are needed and how many grams of CO₂ are formed?

Result: 79.8 g of O₂; 65.9 g of CO₂.

Solution

Balanced equation: $\text{C}_3\text{H}_8 + 5\,\text{O}_2 \to 3\,\text{CO}_2 + 4\,\text{H}_2\text{O}$ (3 C on the right in CO₂, 8 H in 4 H₂O, hence $6 + 4 = 10$ O atoms, i.e. 5 O₂).

Moles of propane: $M = 3\cdot 12.01 + 8\cdot 1.008 = 44.09$ g/mol, $n = 22.0/44.09 = 0.4990$ mol.

Oxygen: $5\cdot 0.4990 = 2.495$ mol, $2.495\cdot 32.00 = 79.8$ g.

Carbon dioxide: $3\cdot 0.4990 = 1.497$ mol, $1.497\cdot 44.01 = 65.9$ g.

Mass conservation check: $4\cdot 0.4990\cdot 18.02 = 36.0$ g of water also form; reactants $22.0 + 79.8 = 101.8$ g, products $65.9 + 36.0 = 101.9$ g, equal within rounding.

Exercise 4. Empirical and molecular formula

A hydrocarbon contains 85.6 % carbon and 14.4 % hydrogen. Its molar mass is about 56 g/mol. Find the empirical and molecular formulas.

Result: CH₂; C₄H₈.

Solution

In 100 g: C $85.6/12.01 = 7.127$ mol; H $14.4/1.008 = 14.29$ mol.

Dividing by the smallest: C 1; H $14.29/7.127 = 2.005 \approx 2$ . Empirical formula CH₂, with mass $12.01 + 2.016 = 14.03$ g/mol.

Ratio of molar mass to empirical formula mass: $56/14.03 \approx 4$ . Molecular formula C₄H₈ (for example a butene).

Exercise 5. A gas from a decomposition

Heating 12.25 g of potassium chlorate, KClO₃, gives KCl and oxygen. How many litres of O₂ are evolved at 0 °C and 1 atm?

Result: 3.36 L.

Solution

Equation: $2\,\text{KClO}_3 \to 2\,\text{KCl} + 3\,\text{O}_2$ .

$M(\text{KClO}_3) = 39.10 + 35.45 + 48.00 = 122.55$ g/mol, $n = 12.25/122.55 = 0.099\,96$ mol.

Oxygen: $\frac{3}{2}\cdot 0.099\,96 = 0.1499$ mol.

Volume: $V = nRT/P = 0.1499\cdot 0.082\,06\cdot 273.15/1 = 3.36$ L (i.e. 0.1499 mol times 22.41 L/mol).

Intermediate level

Exercise 6. Limiting reagent and excess reagent

10.0 g of aluminium react with 30.0 g of chlorine, according to $2\,\text{Al} + 3\,\text{Cl}_2 \to 2\,\text{AlCl}_3$ . How much aluminium chloride forms? Which reagent is left over, and how much?

Moles before and after the reaction between aluminium and chlorine — On the left the initial moles, on the right the final ones. Chlorine is more abundant in moles but is the limiting reagent, because 3 are needed for every 2 of aluminium: at the end it is used up and some aluminium is left over.

Result: 37.6 g of AlCl₃; 2.39 g of Al left over.

Solution

Moles: Al $10.0/26.98 = 0.3706$ mol; Cl₂ $30.0/70.90 = 0.4231$ mol.

Ratios to the coefficients: Al $0.3706/2 = 0.1853$ ; Cl₂ $0.4231/3 = 0.1410$ . The smaller is chlorine’s: chlorine is limiting, even though in moles it is more abundant than aluminium.

Maximum extent $\xi = 0.1410$ mol. AlCl₃: $2\xi = 0.2821$ mol; $M = 26.98 + 3\cdot 35.45 = 133.33$ g/mol; mass $0.2821\cdot 133.33 = 37.6$ g.

Aluminium consumed: $2\xi = 0.2821$ mol; left over $0.3706 - 0.2821 = 0.0885$ mol, i.e. $0.0885\cdot 26.98 = 2.39$ g.

Check: $37.6 + 2.39 = 40.0$ g, equal to the initial mass of the reagents.

Exercise 7. Purity and yield

200 g of a limestone containing 85.0 % CaCO₃ by mass are decomposed by heating: $\text{CaCO}_3 \to \text{CaO} + \text{CO}_2$ . 70.0 g of CO₂ are collected. What is the percentage yield?

Result: 93.6 %.

Solution

Pure carbonate: $0.850\cdot 200 = 170$ g, i.e. $170/100.09 = 1.698$ mol.

Theoretical CO₂: 1.698 mol (1 to 1 ratio), $1.698\cdot 44.01 = 74.7$ g.

Yield: $70.0/74.7 = 93.6$ %. The impurities produce no CO₂ and must be excluded before computing the theoretical yield.

Exercise 8. An acid-base titration

25.0 mL of a sulfuric acid solution are neutralised by 32.4 mL of 0.150 M NaOH. What is the molarity of the acid?

Titration curve of a strong acid with a strong base — The pH stays low while acid is in excess, then rises sharply at the equivalence point, where the moles of base added are twice those of sulfuric acid, and finally levels off with the base in excess.

Result: 0.0972 M.

Solution

Reaction: $\text{H}_2\text{SO}_4 + 2\,\text{NaOH} \to \text{Na}_2\text{SO}_4 + 2\,\text{H}_2\text{O}$ : the acid is diprotic, one mole of acid consumes two moles of base.

Moles of NaOH: $0.150\cdot 0.0324 = 4.86\cdot 10^{-3}$ mol. Moles of acid: $2.43\cdot 10^{-3}$ mol.

Molarity: $2.43\cdot 10^{-3}/0.0250 = 0.0972$ M. The typical mistake is forgetting the factor 2 and finding double.

Exercise 9. Combustion analysis

0.500 g of a compound containing C, H and O burns completely giving 0.999 g of CO₂ and 0.409 g of H₂O. The molar mass of the compound is 88.1 g/mol. Find the molecular formula.

Result: C₄H₈O₂.

Solution

All the carbon ends up in CO₂: $0.999/44.01 = 0.022\,70$ mol of C, i.e. $0.022\,70\cdot 12.01 = 0.2726$ g.

All the hydrogen ends up in water: $0.409/18.02 = 0.022\,70$ mol of H₂O, hence $0.045\,39$ mol of H, i.e. 0.0458 g.

The oxygen of the compound is found by difference (not from the products, because combustion adds oxygen): $0.500 - 0.2726 - 0.0458 = 0.1816$ g, i.e. $0.011\,35$ mol.

Ratios, dividing by 0.011 35: C 2.00; H 4.00; O 1.00. Empirical formula C₂H₄O, mass 44.05 g/mol. $88.1/44.05 = 2$ : molecular formula C₄H₈O₂ (for example ethyl acetate).

Exercise 10. A gas at non-standard conditions

1.20 g of magnesium react with excess hydrochloric acid: $\text{Mg} + 2\,\text{HCl} \to \text{MgCl}_2 + \text{H}_2$ . What volume of hydrogen is collected at 25 °C and 0.980 atm?

Result: 1.23 L.

Solution

Moles of Mg: $1.20/24.31 = 0.049\,36$ mol, equal to the moles of H₂.

$V = nRT/P = 0.049\,36\cdot 0.082\,06\cdot 298.15/0.980 = 1.23$ L. 22.4 L/mol cannot be used: it holds only at 0 °C and 1 atm.

Exam level

Exercise 11. Redox titration of an iron ore

1.000 g of an iron ore is dissolved in acid and all the iron brought to Fe²⁺. The solution requires 22.5 mL of 0.0200 M KMnO₄ to be titrated in acidic solution. Compute the percentage by mass of iron in the ore.

Result: 12.6 %.

Solution

Reaction balanced with half-reactions (in-depth study): reduction $\text{MnO}_4^- + 8\,\text{H}^+ + 5\,e^- \to \text{Mn}^{2+} + 4\,\text{H}_2\text{O}$ ; oxidation $\text{Fe}^{2+} \to \text{Fe}^{3+} + e^-$ , to be multiplied by 5:

$\text{MnO}_4^- + 5\,\text{Fe}^{2+} + 8\,\text{H}^+ \to \text{Mn}^{2+} + 5\,\text{Fe}^{3+} + 4\,\text{H}_2\text{O}$

Moles of permanganate: $0.0200\cdot 0.0225 = 4.50\cdot 10^{-4}$ mol. Moles of iron: $5\cdot 4.50\cdot 10^{-4} = 2.25\cdot 10^{-3}$ mol, i.e. $2.25\cdot 10^{-3}\cdot 55.85 = 0.1257$ g.

Percentage: $0.1257/1.000 = 12.6$ %.

Exercise 12. Analysis of a mixture

A 2.00 g sample is a mixture of NaCl and KCl. Dissolved in water and treated with excess AgNO₃, it gives 4.30 g of AgCl. Compute the percentage by mass of NaCl in the mixture.

Mass of silver chloride as a function of the mixture composition — For a 2.00 g sample the mass of precipitate varies linearly between the value for pure KCl and that for pure NaCl. The measured mass picks out the fraction of NaCl on the line: the method is more accurate the steeper the line, that is, the more different the molar masses.

Result: 42.9 %.

Solution

Both chlorides give one mole of AgCl per mole ( $\text{Cl}^- + \text{Ag}^+ \to \text{AgCl}$ ). Total moles of chloride: $4.30/143.32 = 0.030\,00$ mol ( $M(\text{AgCl}) = 107.87 + 35.45 = 143.32$ ).

Let $x$ be the mass of NaCl ( $M = 58.44$ ); that of KCl ( $M = 74.55$ ) is $2.00 - x$ :

$\frac{x}{58.44} + \frac{2.00 - x}{74.55} = 0.030\,00$

$0.017\,112\,x + 0.026\,828 - 0.013\,414\,x = 0.030\,00$ , i.e. $0.003\,698\,x = 0.003\,175$ , $x = 0.859$ g.

Percentage of NaCl: $0.859/2.00 = 42.9$ %. The system works because the two molar masses are different (in-depth study): the same mass of NaCl contains more chloride than one of KCl.

Exercise 13. A multi-stage industrial process

Nitric acid is produced from ammonia in three stages (Ostwald process):

$4\,\text{NH}_3 + 5\,\text{O}_2 \to 4\,\text{NO} + 6\,\text{H}_2\text{O}$ , yield 90.0 %;
$2\,\text{NO} + \text{O}_2 \to 2\,\text{NO}_2$ , yield 95.0 %;
$3\,\text{NO}_2 + \text{H}_2\text{O} \to 2\,\text{HNO}_3 + \text{NO}$ , yield 100 %.

How many kg of HNO₃ are obtained from 100 kg of NH₃, without recycling the NO produced in the third stage?

Multi-stage process from ammonia to nitric acid — Each stage starts from what the previous one produced, so the overall yield is the product of the yields of the individual stages.

Result: about 211 kg.

Solution

Moles of ammonia: $100,000/17.03 = 5872$ mol.

Stage 1: theoretical NO 5872 mol (4 to 4 ratio), actual $0.900\cdot 5872 = 5285$ mol.

Stage 2: NO₂ $0.950\cdot 5285 = 5021$ mol (2 to 2 ratio).

Stage 3: HNO₃ $\frac{2}{3}\cdot 5021 = 3347$ mol, i.e. $3347\cdot 63.01 = 210.9$ kg.

Remark: the overall nitrogen yield is $0.900\cdot 0.950\cdot\frac{2}{3} = 57$ %. In real plants the NO from the third stage is re-oxidised and recycled, and the yield rises above 90 %.

Exercise 14. A gas mixture

10.0 L of a mixture of methane and ethane, burned completely, consume 26.0 L of oxygen, measured at the same temperature and pressure. Find the volume composition of the mixture and the volume of CO₂ produced.

Result: 60 % CH₄, 40 % C₂H₆; 14.0 L of CO₂.

Solution

At the same temperature and pressure, gas volumes are proportional to moles (Avogadro’s law), so one can reason directly in litres.

Reactions: $\text{CH}_4 + 2\,\text{O}_2 \to \text{CO}_2 + 2\,\text{H}_2\text{O}$ and $\text{C}_2\text{H}_6 + \frac{7}{2}\,\text{O}_2 \to 2\,\text{CO}_2 + 3\,\text{H}_2\text{O}$ .

With $x$ litres of methane and $y$ of ethane: $x + y = 10.0$ and $2x + 3.5y = 26.0$ . Substituting $y = 10.0 - x$ : $2x + 35.0 - 3.5x = 26.0$ , hence $x = 6.0$ L and $y = 4.0$ L.

CO₂: $6.0 + 2\cdot 4.0 = 14.0$ L.

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Basic level

Exercise 1. Moles, formula units and atoms

Exercise 2. Percentage composition

Exercise 3. Combustion of propane

Exercise 4. Empirical and molecular formula

Exercise 5. A gas from a decomposition

Intermediate level

Exercise 6. Limiting reagent and excess reagent

Exercise 7. Purity and yield

Exercise 8. An acid-base titration

Exercise 9. Combustion analysis

Exercise 10. A gas at non-standard conditions

Exam level

Exercise 11. Redox titration of an iron ore

Exercise 12. Analysis of a mixture

Exercise 13. A multi-stage industrial process

Exercise 14. A gas mixture

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