Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
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In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
R2.1.1 Chemical equation
Chemical equation
A symbolic representation of a reaction in which the formulae of the reactants are written on the left and the products on the right. A balanced equation has the same number of atoms of each element on both sides (mass is conserved). The coefficients show the ratio of the amounts (in mol) of reactants and products, e.g. 2H₂(g) + O₂(g) → 2H₂O(l): 2 mol H₂ react with 1 mol O₂ to form 2 mol H₂O. Equations are balanced only by changing coefficients; the formulae (subscripts) are fixed by the substances themselves.
State symbols
Letters in brackets after each formula that show the physical state of that substance under the conditions of the reaction: (s) solid, (l) liquid, (g) gas and (aq) aqueous, meaning dissolved in water. A precipitate is (s), a gas given off is (g), water formed at room temperature is (l), and a substance in solution is (aq), e.g. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g).
Coefficient (stoichiometric coefficient)
The number written in front of a formula in a balanced equation. It multiplies the whole formula: 3O₂ means three O₂ molecules (six O atoms), or 3 mol of O₂. Coefficients give ratios of amounts of substance, not ratios of masses.
Students often think An equation can be balanced by changing the subscripts in a formula, e.g. writing AlO₂ instead of Al₂O₃ so that the oxygen atoms match. In fact No. Only the coefficients may be changed. The formula of each substance is fixed by its composition (for an ionic compound, by the charges of its ions), so aluminium oxide is always Al₂O₃.
Students often think Elements are written as single atoms in equations, so oxygen is written O, e.g. 2Al + 3O → Al₂O₃. In fact As O₂. Oxygen gas consists of diatomic molecules, as do H₂, N₂, F₂, Cl₂, Br₂ and I₂, so these elements are written with the subscript 2.
R2.1.2 Mole ratio
Mole ratio
The ratio of the amounts (in mol) of any two species in a reaction, read from the coefficients of the balanced equation. For Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g), n(Fe₂O₃) : n(Fe) = 1 : 2. It is used to convert the amount of one species into the amount of another.
Relative atomic mass, A_r, and relative formula mass, M_r
A_r is the weighted mean mass of the atoms of an element relative to one-twelfth of the mass of a carbon-12 atom; M_r is the sum of the A_r values of the atoms in a formula, e.g. M_r(Fe₂O₃) = 2(55.85) + 3(16.00) = 159.70. Both are dimensionless; in calculations A_r values are used to two decimal places. Numerically, the molar mass M in g mol⁻¹ equals M_r.
Amount of substance from mass
n = m ÷ M, where n is the amount of substance in mol, m the mass in g and M the molar mass in g mol⁻¹ of that same species. In reacting-mass problems: mass of the given species → amount (n = m/M) → amount of the wanted species (mole ratio) → mass of the wanted species (m = nM).
Molar volume of a gas
The volume occupied by 1 mol of any (ideal) gas at a stated temperature and pressure, V_m. At STP (273 K, 100 kPa) V_m = 22.7 dm³ mol⁻¹, so V = n × V_m. Because equal volumes of gases at the same temperature and pressure contain equal amounts, the volumes of reacting gases are in the same ratio as their coefficients.
Molar concentration
The amount of solute per unit volume of solution, c = n ÷ V, with units mol dm⁻³ when V is in dm³. Volumes measured in cm³ must be divided by 1000 (1 dm³ = 1000 cm³) before use. In reactions in solution, n = cV of one species, the mole ratio and c = n/V of another give an unknown concentration or volume.
Students often think The amount (in mol) of the wanted substance is the same as the amount of the given substance, so the mole ratio in the equation can be ignored. In fact Only if their coefficients are equal. Otherwise the amount must be scaled by the mole ratio from the balanced equation.
Students often think When converting between mass and amount, the M_r of a substance is multiplied by its coefficient in the equation, e.g. using 2 × 84.01 for 2NaHCO₃ or 2 × 55.85 for 2Fe. In fact No. n = m/M uses the molar mass of one formula unit. The coefficient is used only once, in the mole ratio step.
R2.1.3 Limiting reactant
Limiting reactant
The reactant that is completely used up when a reaction goes to completion, and so determines the maximum amount of product. It is found by comparing the amounts present with the mole ratio in the equation (e.g. dividing each amount by its coefficient), not by comparing masses, and not by comparing amounts without the ratio.
Excess reactant
A reactant present in more than the amount needed to react with all of the limiting reactant. Some of it remains unreacted at the end; the amount left = amount present − amount that reacted with the limiting reactant.
Theoretical yield
The maximum mass (or amount) of product that could be obtained, calculated from the amount of the limiting reactant using the mole ratio, assuming the reaction goes to completion with no losses.
Experimental yield
The mass (or amount) of product actually obtained in a practical preparation. It is normally less than the theoretical yield because of incomplete reaction, side reactions and losses during transfer, filtration and purification; an apparent yield above the theoretical yield indicates an impure (e.g. wet) product or an error.
Students often think The reactant with the smaller mass runs out first, so it is the limiting reactant. In fact No. Reactants react in fixed ratios of amounts, not masses. The limiting reactant is found by converting masses to amounts and comparing them with the mole ratio.
Students often think The reactant present in the smaller amount (in mol) is the limiting reactant, whatever the coefficients in the equation. In fact Only when the coefficients are equal. For 2Mg + O₂ → 2MgO, 0.200 mol Mg needs just 0.100 mol O₂, so with 0.125 mol O₂ present magnesium is limiting even though there is more of it in mol.
R2.1.4 Percentage yield
Percentage yield
Percentage yield = (experimental yield ÷ theoretical yield) × 100%, with both yields in the same units (mass or amount) and for the same product. It measures how much of the maximum possible product was actually obtained.
Students often think The theoretical yield is the typical or expected yield, so a reaction that runs particularly efficiently can give more product than the theoretical yield. In fact No. The theoretical yield is the maximum possible mass of product that the limiting reactant can form. A percentage yield above 100% means that the product weighed is impure (e.g. wet) or that there has been an error.
Students often think A catalyst increases the yield of a reaction, so it can raise the mass of product above what would otherwise be possible. In fact No. A catalyst increases the rate of reaction; it does not change the amount of limiting reactant, so the theoretical yield is unchanged.
R2.1.5 Atom economy
Atom economy
A measure of the efficiency of a reaction in green chemistry: atom economy = [M_r of desired product ÷ sum of M_r of all reactants] × 100%, using the coefficients of the balanced equation (e.g. for C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂, the numerator is 2 × M_r(C₂H₅OH)). It depends only on the stoichiometry, not on the masses used or the yield obtained.
Atom economy and waste
Atom economy and waste are inversely related: the lower the atom economy, the greater the proportion of the mass of the reactants that ends up in by-products rather than the desired product, and so the more waste an industrial process generates. Addition reactions with a single product have an atom economy of 100%. Atom economy does not account for yield, unreacted reactants, solvents or energy.
Students often think Atom economy and percentage yield measure the same thing, so increasing the percentage yield raises the atom economy. In fact No. Atom economy is calculated from the equation alone and is fixed for a given reaction; percentage yield depends on how the experiment is carried out. A reaction can have a high yield but a low atom economy, or the reverse.
Students often think Atom economy is calculated from the actual masses of product and reactants in a batch, so it changes with the quantities used. In fact No. Atom economy is calculated from the M_r values and coefficients in the balanced equation, so it is the same whatever masses are used.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Hydrogen burns in oxygen, and the quantities are measured at room temperature and pressure: 2H₂(g) + O₂(g) → 2H₂O(l). Which statement about the quantities shown by this equation is correct?
Answer and reasoning
The mass of H₂ that reacts is twice the mass of O₂ that reacts. — A student who reads the coefficients as a mass ratio picks this. The ratio 2 : 1 is by amount. In mass, 2 mol H₂ is 4.04 g and 1 mol O₂ is 32.00 g, so far less hydrogen than oxygen reacts by mass.
Every 2 mol of H₂ react with 1 mol of O₂ to form 2 mol of H₂O. — The coefficients give the ratio of amounts in mol: 2 mol H₂ : 1 mol O₂ : 2 mol H₂O. Mass is conserved (4.04 g + 32.00 g = 36.04 g), but the amounts and masses of the individual substances are not in the same ratio.
Every 2 dm³ of H₂ and 1 dm³ of O₂ react to form 2 dm³ of H₂O. — A student who applies gas-volume ratios to every substance picks this. The 2 : 1 volume ratio holds for the two gases, but water forms as a liquid, H₂O(l); 2 mol of liquid water occupies only about 36 cm³.
The amount of H₂O formed equals the total amount of H₂ and O₂ used. — A student who thinks that amount in mol is conserved, like mass, picks this. Atoms are conserved, but here 3 mol of reactant molecules form 2 mol of water molecules.
2 Iron is produced by reducing iron(III) oxide with carbon monoxide: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g). What is the maximum mass of iron that can be produced from 40.0 g of iron(III) oxide? A_r: O 16.00; Fe 55.85
Answer and reasoning
28.0 g — M_r(Fe₂O₃) = 2(55.85) + 3(16.00) = 159.70, so n(Fe₂O₃) = 40.0 ÷ 159.70 = 0.2505 mol. The ratio Fe₂O₃ : Fe is 1 : 2, so n(Fe) = 0.5009 mol, and m(Fe) = 0.5009 × 55.85 = 28.0 g.
6.99 g — A student who inverts the mole ratio divides by 2 instead of multiplying: n(Fe) = 0.2505 ÷ 2 = 0.1252 mol, and 0.1252 × 55.85 = 6.99 g. Each Fe₂O₃ contains two Fe atoms, so there must be more Fe (in mol) than Fe₂O₃, not less.
56.0 g — A student who multiplies the A_r of iron by its coefficient finds the correct n(Fe) = 0.5009 mol but then uses 2 × 55.85 = 111.70 g mol⁻¹, giving 56.0 g. The coefficient has already been used in the mole ratio; m = nM uses the molar mass of one Fe atom, 55.85 g mol⁻¹.
80.0 g — A student who reads the coefficients as a mass ratio doubles the mass of iron(III) oxide: 2 × 40.0 = 80.0 g. That would be more iron than the oxide contains; the 1 : 2 ratio applies to amounts in mol, not masses.
Working M_r(Fe₂O₃) = 2(55.85) + 3(16.00) = 159.70. n(Fe₂O₃) = 40.0 g ÷ 159.70 g mol⁻¹ = 0.2505 mol. Mole ratio Fe₂O₃ : Fe = 1 : 2, so n(Fe) = 2 × 0.2505 = 0.5009 mol. m(Fe) = 0.5009 mol × 55.85 g mol⁻¹ = 27.98 g ≈ 28.0 g.
3 Which statement describes the theoretical yield of a product in a reaction?
Answer and reasoning
The mass of product actually obtained at the end of the experiment — A student who confuses the two yields picks this. The mass actually obtained is the experimental yield; the theoretical yield is calculated, and the experimental yield is normally smaller.
The total mass of all the reactants put in, since mass is conserved — A student who thinks that all the reactants turn into product picks this. Only the limiting reactant is used up and some excess reactant is left over; other products may also form, so the product mass is not the total mass of the reactants.
The mass of product calculated from the amount of the limiting reactant — The theoretical yield is the maximum mass of product, calculated from the amount of the limiting reactant using the mole ratio and assuming complete reaction with no losses.
The mass of product calculated from the reactant with the smaller number of moles — A student who compares amounts without the mole ratio picks this. The limiting reactant is not always the one present in the smaller amount: for 2Mg + O₂ → 2MgO, 0.200 mol Mg with 0.125 mol O₂ leaves magnesium limiting.
4 A student heats 2.70 g of aluminium in 8.51 g of chlorine and collects 7.60 g of aluminium chloride: 2Al(s) + 3Cl₂(g) → 2AlCl₃(s). What is the percentage yield of aluminium chloride? A_r: Al 26.98; Cl 35.45
Answer and reasoning
57.0% — A student who takes the reactant with fewer moles as limiting chooses aluminium (0.1001 mol < 0.1200 mol) and gets a theoretical yield of 0.1001 × 133.33 = 13.34 g, so 7.60 ÷ 13.34 × 100 = 57.0%. Aluminium needs 1.5 mol Cl₂ per mol, so chlorine runs out first.
67.8% — A student who assumes that all of the reactants turn into product takes the theoretical yield as 2.70 + 8.51 = 11.21 g, giving 67.8%. Aluminium is in excess and some is left over, so the maximum mass of product is 10.67 g.
89.3% — A student who compares the product with the mass of the limiting reactant divides 7.60 g by the 8.51 g of chlorine, giving 89.3%. The theoretical yield is a mass of aluminium chloride, which also contains aluminium: 0.08002 mol × 133.33 g mol⁻¹ = 10.67 g.
71.2% — n(Al) = 0.1001 mol and n(Cl₂) = 0.1200 mol. Al would need 1.5 × 0.1001 = 0.1501 mol Cl₂, so Cl₂ is limiting. n(AlCl₃) = 0.1200 × 2/3 = 0.08002 mol, theoretical yield = 0.08002 × 133.33 = 10.67 g, and percentage yield = 7.60 ÷ 10.67 × 100 = 71.2%.
Working n(Al) = 2.70 g ÷ 26.98 g mol⁻¹ = 0.1001 mol; n(Cl₂) = 8.51 g ÷ 70.90 g mol⁻¹ = 0.1200 mol. Al would need 3/2 × 0.1001 = 0.1501 mol Cl₂, but only 0.1200 mol is present, so Cl₂ is limiting. n(AlCl₃) = 2/3 × 0.1200 = 0.08002 mol. M_r(AlCl₃) = 26.98 + 3(35.45) = 133.33. Theoretical yield = 0.08002 mol × 133.33 g mol⁻¹ = 10.67 g. Percentage yield = 7.60 g ÷ 10.67 g × 100 = 71.2%.
5 Which statement about atom economy and waste in an industrial process is correct?
Answer and reasoning
The higher the percentage yield, the higher the atom economy, so less waste is produced. — A student who confuses atom economy with percentage yield picks this. Atom economy is fixed by the balanced equation; improving the yield does not change the proportion of the reactant mass that goes into by-products.
Atom economy counts one molecule of each substance, so the coefficients do not affect it. — A student who leaves the coefficients out of the calculation picks this. The masses must match the balanced equation: coefficient × M_r for the desired product and for each reactant, so changing the coefficients changes the atom economy and the proportion of mass that becomes waste.
A process with an atom economy of 100% produces no waste of any kind at all. — A student who treats atom economy as a complete measure of waste picks this. An atom economy of 100% means that the reaction forms no by-product, but unreacted reactants, side reactions and solvents can still create waste.
The lower the atom economy, the greater the proportion of reactant mass that becomes waste. — Atom economy is the percentage of the total mass of reactants that can end up in the desired product. The rest ends up in by-products, so as atom economy falls, the proportion of the reactant mass converted to waste rises: the two are inversely related.
6 Aluminium burns in oxygen to form aluminium oxide. Which equation for this reaction is correctly balanced?
Answer and reasoning
Al(s) + O₂(g) → AlO₂(s) — A student who balances by changing a formula picks this. The atoms match, but AlO₂ is not aluminium oxide: the charges Al³⁺ and O²⁻ fix the formula as Al₂O₃. Only coefficients may be changed when balancing.
2Al(s) + 3O(g) → Al₂O₃(s) — A student who writes oxygen as single atoms picks this. The atom counts match, but oxygen exists as O₂ molecules, so the equation must use O₂; with O₂ the balanced equation is 4Al(s) + 3O₂(g) → 2Al₂O₃(s).
2Al(s) + 3O₂(g) → Al₂O₃(s) — A student who reads 3O₂ as three oxygen atoms picks this. The coefficient multiplies the whole formula: 3O₂ contains 3 × 2 = 6 O atoms but Al₂O₃ contains only 3, so oxygen is not balanced.
4Al(s) + 3O₂(g) → 2Al₂O₃(s) — Aluminium oxide is Al₂O₃ (ions Al³⁺ and O²⁻) and oxygen gas is O₂. 3O₂ supplies 6 O atoms, matching 2Al₂O₃, which contains 4 Al atoms, so 4Al. Check: 4 Al and 6 O on each side.
7 Sodium hydrogencarbonate decomposes on heating: 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g). What volume of carbon dioxide, measured at STP, is produced when 16.8 g of sodium hydrogencarbonate decomposes completely? A_r: H 1.01; C 12.01; O 16.00; Na 22.99. Molar volume of an ideal gas at STP = 22.7 dm³ mol⁻¹
Answer and reasoning
4.54 dm³ — A student who ignores the mole ratio takes n(CO₂) = n(NaHCO₃) = 0.200 mol and gets 0.200 × 22.7 = 4.54 dm³. Two NaHCO₃ produce only one CO₂, so the amount must be halved.
2.27 dm³ — M_r(NaHCO₃) = 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01, so n(NaHCO₃) = 16.8 ÷ 84.01 = 0.200 mol. The ratio NaHCO₃ : CO₂ is 2 : 1, so n(CO₂) = 0.100 mol and V = 0.100 × 22.7 = 2.27 dm³.
4.33 dm³ — A student who reads the 2 : 1 coefficients as a mass ratio takes the mass of CO₂ as 16.8 ÷ 2 = 8.40 g, so n(CO₂) = 8.40 ÷ 44.01 = 0.191 mol and V = 0.191 × 22.7 = 4.33 dm³. The ratio applies to amounts: n(NaHCO₃) = 0.200 mol gives n(CO₂) = 0.100 mol.
8.67 dm³ — A student who divides the mass of NaHCO₃ by the M_r of CO₂ gets 16.8 ÷ 44.01 = 0.382 mol and 0.382 × 22.7 = 8.67 dm³. The 16.8 g is sodium hydrogencarbonate, so it must be divided by M_r(NaHCO₃) before the mole ratio is used.
Working M_r(NaHCO₃) = 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01. n(NaHCO₃) = 16.8 g ÷ 84.01 g mol⁻¹ = 0.2000 mol. Mole ratio NaHCO₃ : CO₂ = 2 : 1, so n(CO₂) = 0.1000 mol. V(CO₂) = n × V_m = 0.1000 mol × 22.7 dm³ mol⁻¹ = 2.27 dm³.
8 A student heats 4.86 g of magnesium with 4.00 g of oxygen: 2Mg(s) + O₂(g) → 2MgO(s). Which statement about the limiting reactant is correct? A_r: O 16.00; Mg 24.31
Answer and reasoning
Magnesium is limiting: 0.200 mol Mg needs 0.100 mol O₂, and 0.125 mol is present. — n(Mg) = 4.86 ÷ 24.31 = 0.200 mol and n(O₂) = 4.00 ÷ 32.00 = 0.125 mol. The 2 : 1 ratio means 0.200 mol Mg needs only 0.100 mol O₂, so oxygen is in excess (0.025 mol left) and magnesium is limiting.
Oxygen is limiting, because a smaller mass of oxygen than of magnesium is present. — A student who compares masses picks this. Reactants react in fixed ratios of amounts: 4.00 g O₂ is 0.125 mol, more than the 0.100 mol that 0.200 mol Mg needs, so oxygen is in excess.
Oxygen is limiting, because there is only 0.125 mol of O₂ but 0.200 mol of Mg present. — A student who compares amounts without the mole ratio picks this. The equation shows 2 mol Mg react with 1 mol O₂, so 0.200 mol Mg needs only 0.100 mol O₂; with 0.125 mol O₂ present, oxygen is in excess.
Magnesium is limiting: 4.86 g Mg needs only 2.43 g O₂, and 4.00 g is present. — A student who treats the 2 : 1 coefficients as a mass ratio halves the mass of magnesium to get 2.43 g. The ratio is by amount: 0.200 mol Mg needs 0.100 mol O₂, which is 3.20 g. The conclusion happens to be right, but the reasoning and the 2.43 g are wrong.
Working n(Mg) = 4.86 g ÷ 24.31 g mol⁻¹ = 0.200 mol; n(O₂) = 4.00 g ÷ 32.00 g mol⁻¹ = 0.125 mol. O₂ needed for 0.200 mol Mg = 0.200 ÷ 2 = 0.100 mol (3.20 g). 0.125 mol O₂ is present, so O₂ is in excess (0.025 mol, 0.800 g, left over) and Mg is the limiting reactant.
9 A student prepares a solid product, filters it off and weighs it. The theoretical yield was calculated correctly from the limiting reactant, but the percentage yield comes out as 108%. Which is the most likely explanation?
Answer and reasoning
The product was weighed before it had been completely dried. — The theoretical yield is the maximum mass of product the limiting reactant can form, so a yield above 100% means that the solid weighed was not pure product. Water left in an undried solid adds mass and is the usual cause.
A catalyst was present, and it increased the mass of product formed. — A student who thinks catalysts increase yield picks this. A catalyst only speeds up the reaction; the maximum mass of product is still fixed by the amount of the limiting reactant.
The excess reactant formed more product after the limiting one ran out. — A student who thinks all the reactants turn into product picks this. Every unit of product needs the limiting reactant; once that is used up, the excess reactant cannot form any more product and is simply left over.
The reaction ran more efficiently than usual and beat its expected yield. — A student who treats the theoretical yield as a typical figure that can be beaten picks this. It is the maximum set by the atoms in the limiting reactant; no reaction, however efficient, can exceed it.
10 Hydrogen is manufactured by reacting methane with steam: CH₄(g) + H₂O(g) → CO(g) + 3H₂(g). In one batch, 16.0 kg of methane gave 4.80 kg of hydrogen. What is the atom economy for producing hydrogen by this reaction? A_r: H 1.01; C 12.01; O 16.00
Answer and reasoning
5.93% — A student who leaves out the coefficient uses one H₂: 2.02 ÷ 34.07 × 100 = 5.93%. Each CH₄ gives three H₂ molecules, so the desired product counts as 3 × 2.02 = 6.06.
30.0% — A student who uses the batch masses divides 4.80 kg by 16.0 kg to get 30.0%. Atom economy uses M_r values and coefficients from the balanced equation, so it does not depend on the masses used in a particular batch.
17.8% — Atom economy = M_r(desired product) ÷ ΣM_r(all reactants) × 100, using the coefficients: 3 × 2.02 ÷ (16.05 + 18.02) × 100 = 6.06 ÷ 34.07 × 100 = 17.8%. The rest of the mass ends up as the by-product CO.
37.8% — A student who leaves the steam out of the denominator uses M_r(CH₄) alone: 6.06 ÷ 16.05 × 100 = 37.8%. Every reactant in the equation counts: 16.05 + 18.02 = 34.07.
Working M_r(CH₄) = 12.01 + 4(1.01) = 16.05; M_r(H₂O) = 2(1.01) + 16.00 = 18.02; M_r(H₂) = 2(1.01) = 2.02. Atom economy = [3 × M_r(H₂) ÷ (M_r(CH₄) + M_r(H₂O))] × 100 = (6.06 ÷ 34.07) × 100 = 17.8%. The batch masses are not needed: atom economy depends only on the stoichiometry of the equation.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
4 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Each reaction below is carried out in the laboratory by adding the first reactant (a solid, or potassium hydroxide solution) to a dilute aqueous acid. Which equation shows correct state symbols?
Answer and reasoning
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(aq) — A student who uses (aq) for everything that forms in the solution picks this. (aq) means dissolved in water; hydrogen bubbles out of the mixture as a gas, so it is H₂(g).
KOH(aq) + HCl(aq) → KCl(l) + H₂O(l) — A student who thinks a solution is a liquid gives the dissolved potassium chloride the symbol (l). It is dissolved in water, so it is KCl(aq); only the water formed is a liquid in its own right.
MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) — Magnesium oxide is added as a solid (s); the dilute acid and the magnesium chloride formed are dissolved in water (aq); the water formed is a liquid (l).
CuO(s) + 2HNO₃(aq) → Cu(NO₃)₂(s) + H₂O(l) — A student who writes the state of the pure substance gives copper(II) nitrate the symbol (s), as it is a solid in the bottle. Here it forms dissolved in the dilute acid, so it is Cu(NO₃)₂(aq).
2 A 0.399 g sample of iron(III) oxide reacts with exactly 30.0 cm³ of hydrochloric acid, with neither reactant left over: Fe₂O₃(s) + 6HCl(aq) → 2FeCl₃(aq) + 3H₂O(l). What is the concentration of the hydrochloric acid? A_r: H 1.01; O 16.00; Cl 35.45; Fe 55.85
Answer and reasoning
2.19 mol dm⁻³ — A student who reads the coefficients as a mass ratio takes the mass of HCl as 6 × 0.399 g = 2.394 g, so n(HCl) = 2.394 ÷ 36.46 = 0.06566 mol and c = 0.06566 ÷ 0.0300 = 2.19 mol dm⁻³. The 1 : 6 ratio applies to amounts, not masses: n(HCl) = 6 × 2.498 × 10⁻³ = 1.499 × 10⁻² mol.
0.000500 mol dm⁻³ — A student who leaves the volume in cm³ divides 1.499 × 10⁻² mol by 30.0 and gets 0.000500. For mol dm⁻³ the volume must be in dm³: 30.0 cm³ = 0.0300 dm³, giving 0.500 mol dm⁻³.
0.0139 mol dm⁻³ — A student who inverts the mole ratio divides by 6 instead of multiplying: n(HCl) = 2.498 × 10⁻³ ÷ 6 = 4.164 × 10⁻⁴ mol, and 4.164 × 10⁻⁴ ÷ 0.0300 = 0.0139 mol dm⁻³. One Fe₂O₃ reacts with six HCl, so there must be more HCl than Fe₂O₃, not less.
0.500 mol dm⁻³ — n(Fe₂O₃) = 0.399 ÷ 159.70 = 2.498 × 10⁻³ mol. The ratio Fe₂O₃ : HCl is 1 : 6, so n(HCl) = 1.499 × 10⁻² mol. With V = 30.0 cm³ = 0.0300 dm³, c = 1.499 × 10⁻² ÷ 0.0300 = 0.500 mol dm⁻³.
3 5.00 cm³ of propane is mixed with 40.0 cm³ of oxygen and ignited, and the propane burns completely: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. The mixture is then cooled to the original room temperature and pressure. What is the total volume of gas present at the end?
Answer and reasoning
50.0 cm³ — A student who counts the water as a gas adds 4 × 5.00 = 20.0 cm³ of steam to get 15.0 + 15.0 + 20.0 = 50.0 cm³. At room temperature the water condenses to a liquid, whose volume is negligible.
30.0 cm³ — 5.00 cm³ C₃H₈ uses 5 × 5.00 = 25.0 cm³ O₂, so 15.0 cm³ O₂ is left over, and it forms 3 × 5.00 = 15.0 cm³ CO₂. At room temperature the water is a liquid. Total gas = 15.0 + 15.0 = 30.0 cm³.
45.0 cm³ — A student who thinks that gas volume is conserved, like mass, adds the starting volumes: 5.00 + 40.0 = 45.0 cm³. The reaction changes the number of gas molecules: 6 volumes of gaseous reactant give 3 volumes of CO₂ plus liquid water.
15.0 cm³ — A student who assumes that all the reactants are used up counts only the 15.0 cm³ of CO₂. Propane is limiting and uses only 25.0 cm³ of the 40.0 cm³ of oxygen, so 15.0 cm³ of O₂ is still present.
Working At the same temperature and pressure, gas volumes are in the same ratio as amounts. O₂ needed = 5 × 5.00 cm³ = 25.0 cm³ < 40.0 cm³, so propane is limiting and O₂ is in excess. O₂ left = 40.0 − 25.0 = 15.0 cm³. CO₂ formed = 3 × 5.00 = 15.0 cm³. H₂O is a liquid at room temperature (negligible volume). Total gas = 15.0 + 15.0 = 30.0 cm³.
4 Ethanol is manufactured either by fermentation, C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g), or by hydration of ethene, C₂H₄(g) + H₂O(g) → C₂H₅OH(g). Which statement about the two routes is correct? A_r: H 1.01; C 12.01; O 16.00
Answer and reasoning
A higher percentage yield in fermentation would raise its atom economy closer to hydration's. — A student who confuses atom economy with percentage yield picks this. Atom economy is fixed by the balanced equation at 51.1% for fermentation; a higher yield gives more ethanol, but each glucose still forms two CO₂ as well.
Fermentation has the lower atom economy, so more of its reactant mass becomes by-product. — Fermentation: 2 × 46.08 ÷ 180.18 × 100 = 51.1%, so about half of the reactant mass ends up in CO₂. Hydration: 46.08 ÷ (28.06 + 18.02) × 100 = 100%, so no by-product forms. The lower the atom economy, the greater the proportion of reactant mass that becomes by-product waste.
Hydration has an atom economy of 100%, so a hydration plant produces no waste of any kind. — A student who treats atom economy as a complete measure of waste picks this. 100% means the reaction itself forms no by-product, but side reactions, losses of catalyst and the energy used by the plant still create waste.
Only about a quarter of the reactant mass in fermentation ends up in ethanol (atom economy 25.6%). — A student who leaves out the coefficient uses one ethanol: 46.08 ÷ 180.18 × 100 = 25.6%. Each glucose gives two ethanol molecules, so the desired product counts as 2 × 46.08 = 92.16, and the atom economy is 51.1%: about half the mass ends up in ethanol.
Working M_r(C₆H₁₂O₆) = 6(12.01) + 12(1.01) + 6(16.00) = 180.18; M_r(C₂H₅OH) = 2(12.01) + 6(1.01) + 16.00 = 46.08; M_r(C₂H₄) = 2(12.01) + 4(1.01) = 28.06; M_r(H₂O) = 18.02. Fermentation: atom economy = 2(46.08) ÷ 180.18 × 100 = 51.1%, so 48.9% of the reactant mass ends up in the by-product CO₂. Hydration: atom economy = 46.08 ÷ (28.06 + 18.02) × 100 = 100%, so the reaction forms no by-product. Omitting the coefficient 2 in fermentation gives 46.08 ÷ 180.18 × 100 = 25.6%.
That was your twenty minutes. Real practice on R2.1 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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