ESAT Chemistry: Quantitative chemistry and redox, set 1

• Answer all questions. No calculator.
• Each question has exactly one correct answer.
• Use the Ar values given in each question.

1. 11 mark

   Which of the following statements about what happens during a chemical reaction is correct?

   1. A New atoms are created during the reaction, to provide the extra atoms needed to build the bonds in the products.
   2. B The atoms already present are rearranged into new substances; the nuclei of those atoms are neither created nor destroyed during the reaction.
   3. C Atoms of one element can be converted into atoms of a different element during a chemical reaction.
   4. D The total number of electrons on each individual atom must stay exactly the same throughout the whole reaction, since electrons are never gained or lost by atoms.
2. 21 mark

   Aluminium forms the ion Al^3+ and sulfate forms the ion SO4^2-. What is the correct formula, including state symbol, for aluminium sulfate dissolved in water?

   1. A AlSO4 (aq)
   2. B Al3(SO4)2 (aq)
   3. C Al2(SO4)3 (s)
   4. D Al2(SO4)3 (aq)
3. 31 mark

   Propane burns completely in oxygen: C3H8 + O2 -> CO2 + H2O. When this equation is balanced using whole-number coefficients, what is the coefficient of O2?

   1. A 5
   2. B 3
   3. C 10
   4. D 7
4. 41 mark

   In the Haber process, nitrogen and hydrogen reach equilibrium: N2(g) + 3H2(g) <=> 2NH3(g). The forward reaction is exothermic. What is the effect on the position of equilibrium of increasing the overall pressure of the system, at constant temperature?

   1. A The equilibrium shifts to the left, forming more N2 and H2, because increasing the pressure always favours the reverse reaction regardless of the number of gas molecules on each side.
   2. B The equilibrium is not affected by the change in pressure, because ammonia is a covalent gas rather than an ionic solid, and pressure only influences reactions involving ionic substances.
   3. C The equilibrium shifts to the right, forming more ammonia, because increasing the pressure favours the side of the equation with fewer moles of gas, and there are fewer gas molecules among the products (2 mol) than the reactants (4 mol).
   4. D The equilibrium shifts to the right, forming more ammonia, because increasing the pressure also increases the temperature of the system, which favours the exothermic forward reaction.
5. 51 mark

   Using Ar: Na = 23, Cl = 35.5, how many moles of sodium chloride are there in 117 g of the pure solid?

   1. A 0.5 mol
   2. B 2 mol
   3. C 20 mol
   4. D 6844.5 mol
6. 61 mark

   Calculate the percentage by mass of magnesium in magnesium oxide, MgO. Use Ar: Mg = 24, O = 16.

   1. A 60%
   2. B 40%
   3. C 66.7%
   4. D 24%
7. 71 mark

   A compound of iron and oxygen contains 70% iron and 30% oxygen by mass. Using Ar: Fe = 56, O = 16, what is its empirical formula?

   1. A FeO
   2. B FeO2
   3. C Fe7O3
   4. D Fe2O3
8. 81 mark

   Nitrogen reacts with hydrogen to form ammonia: N2(g) + 3H2(g) -> 2NH3(g). 28 g of nitrogen is mixed with 3 g of hydrogen and allowed to react to completion. Using Ar: N = 14, H = 1, what is the maximum mass of ammonia that can be formed?

   1. A 34 g
   2. B 12.75 g
   3. C 17 g
   4. D 25.5 g
9. 91 mark

   2.4 g of magnesium reacts exactly with 1.6 g of oxygen gas to form magnesium oxide. Using Ar: Mg = 24, O = 16, which balanced equation is consistent with this data?

   1. A Mg + O -> MgO
   2. B 2Mg + O2 -> 2MgO
   3. C 2Mg + O2 -> 2MgO2
   4. D 2Mg + O2 -> Mg2O
10. 101 mark

    At room temperature and pressure, one mole of any gas occupies 24 dm3. What volume, in dm3, is occupied by 4.4 g of carbon dioxide gas at room temperature and pressure? Use Ar: C = 12, O = 16.

    1. A 2.4 dm3
    2. B 24 dm3
    3. C 0.24 dm3
    4. D 44 dm3
11. 111 mark

    In a titration, 25.0 cm3 of sodium hydroxide solution exactly neutralises 20.0 cm3 of 0.50 mol dm-3 hydrochloric acid: NaOH + HCl -> NaCl + H2O. What is the concentration of the sodium hydroxide solution, in mol dm-3?

    1. A 0.625 mol dm-3
    2. B 0.01 mol dm-3
    3. C 0.50 mol dm-3
    4. D 0.40 mol dm-3
12. 121 mark

    A reaction is predicted to produce 12.5 g of a product. In practice, 10.0 g is actually obtained. What is the percentage yield?

    1. A 125%
    2. B 20%
    3. C 80%
    4. D 2.5%
13. 131 mark

    In the reaction Fe2O3 + 3CO -> 2Fe + 3CO2, which species is reduced, and why?

    1. A CO, because it gains oxygen atoms to become CO2, and gaining any atoms during a reaction always counts as a reduction.
    2. B Fe2O3, because the iron loses oxygen (and each Fe^3+ ion gains electrons) as it is converted into Fe.
    3. C Fe2O3, because the compound as a whole loses mass when it breaks apart into separate iron and oxygen-containing products.
    4. D CO2, because it is the more chemically complex product formed by the end of the reaction, and reduction is defined as the formation of a more complex substance.
14. 141 mark

    Zinc metal reacts with copper sulfate solution: Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s). Given that the sulfate ion, SO4^2-, has an overall charge of -2, what is the oxidation state of copper in CuSO4, and how should the overall reaction be classified?

    1. A Copper is +2 in CuSO4; the reaction is a redox reaction, since zinc is oxidised (0 to +2) while copper is reduced (+2 to 0).
    2. B Copper is -2 in CuSO4, copying the sulfate ion's own charge directly onto copper instead of balancing it against the compound's overall neutral charge.
    3. C Copper is +2 in CuSO4, but the reaction is classified as oxidation-only, because the question only describes zinc's change in oxidation state.
    4. D Copper is +8 in CuSO4, treating each of the four oxygen atoms as independently contributing -2 with no overall charge assigned to the sulfate group, instead of using the sulfate ion's fixed charge of -2.
15. 151 mark

    Chlorine gas reacts with cold, dilute sodium hydroxide: Cl2 + 2NaOH -> NaCl + NaOCl + H2O. In this reaction, chlorine's oxidation state changes from 0 (in Cl2) to -1 (in NaCl) and to +1 (in NaOCl). Which term correctly describes this, and which species acts as both the oxidising and reducing agent?

    1. A This is disproportionation; sodium hydroxide is both the oxidising agent and the reducing agent, because it is the reactant that supplies the hydroxide ions driving the reaction.
    2. B This is a simple redox reaction, not disproportionation, because two different products (NaCl and NaOCl) are formed rather than one.
    3. C This is disproportionation; chlorine (as Cl2) is both the oxidising agent and the reducing agent, since atoms of the same element are simultaneously oxidised and reduced.
    4. D This is disproportionation; water is the reducing agent, since it is produced as a by-product with hydrogen in an oxidation state of +1 rather than the 0 it would have in principle.

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: B

   1. A chemical reaction rearranges existing atoms into new arrangements (new substances); it does not create or destroy any atoms.
   2. In particular, the nucleus of every atom present is unchanged: no protons or neutrons are gained, lost, or converted into a different element's nucleus.
   3. Electrons can move between atoms during bonding or redox reactions (that is exactly what ionic bonding and oxidation/reduction are), so an individual atom's electron count is not fixed; only the nuclei are untouched.
   4. Statement B is the only one that correctly restricts the 'nothing created or destroyed' rule to nuclei, so B is correct.
   • Why not A: Invents new atoms to explain bonding, when in fact all the atoms needed in the products were already present among the reactants; only the arrangement of existing atoms changes.
   • Why not C: Confuses a chemical reaction with a nuclear reaction; turning one element into another would require changing the number of protons in the nucleus, which chemical reactions never do.
   • Why not D: Overgeneralises 'no atoms created or destroyed' into 'no atom's electron count can change'. Ionic bonding and redox reactions both involve individual atoms gaining or losing electrons, even though the total number of electrons among all the atoms combined stays fixed.

2. Question 2Answer: D

   1. Aluminium forms the Al^3+ ion and sulfate is the polyatomic ion SO4^2-.
   2. To balance the charges, the charge on one ion becomes the subscript of the other: 2 Al^3+ ions (total charge +6) balance 3 SO4^2- ions (total charge -6), giving Al2(SO4)3.
   3. The question states the compound is dissolved in water, so it must carry the state symbol for a dissolved substance, (aq), not (s) for a solid.
   4. Putting the formula and the state symbol together gives Al2(SO4)3 (aq), option D.
   • Why not A: Writes a 1:1 ratio of aluminium to sulfate ions, ignoring that the charges (Al^3+ and SO4^2-) must be balanced by using 2 aluminium ions for every 3 sulfate ions, not 1 of each.
   • Why not B: Balances the charges the wrong way round, putting each ion's own charge number as its own subscript (a 3 from aluminium's own 3+ charge) instead of crossing each ion's charge over onto the OTHER ion.
   • Why not C: Gets the correct formula unit but the wrong state symbol; since the compound is dissolved in water it should be labelled (aq), a solution, not (s) for a solid.

3. Question 3Answer: A

   1. Balance carbon first: propane has 3 carbon atoms, so 3 CO2 molecules are needed on the right.
   2. Balance hydrogen next: propane has 8 hydrogen atoms, so 4 H2O molecules are needed (each H2O carries 2 H atoms, and 4 x 2 = 8).
   3. Now count the oxygen atoms needed on the right: 3 CO2 contributes 3 x 2 = 6 oxygen atoms, and 4 H2O contributes 4 x 1 = 4 oxygen atoms, giving 6 + 4 = 10 oxygen atoms in total.
   4. Since O2 supplies its oxygen atoms two at a time, 10 atoms require 10 / 2 = 5 molecules of O2, so the coefficient of O2 is 5 and the balanced equation is C3H8 + 5O2 -> 3CO2 + 4H2O.
   • Why not B: Matches the coefficient later needed for CO2 (3), from assuming all the product's oxygen comes only from the carbon dioxide and forgetting that the water molecules also need oxygen supplied by O2.
   • Why not C: Correctly counts that 10 oxygen atoms are needed in total (6 in 3CO2 plus 4 in 4H2O) but forgets that O2 is a diatomic molecule, so this atom count must be divided by 2 to get the number of O2 molecules.
   • Why not D: Adds the coefficients needed for carbon and hydrogen (3 + 4 = 7) instead of counting the oxygen atoms required, mixing up unrelated parts of the equation.

4. Question 4Answer: C

   1. The forward reaction, N2(g) + 3H2(g) -> 2NH3(g), has 1 + 3 = 4 moles of gas on the reactant side and 2 moles of gas on the product side.
   2. Le Chatelier's principle says that increasing the overall pressure shifts an equilibrium towards the side with fewer moles of gas, because that reduces the total number of gas particles and partially counteracts the pressure increase.
   3. Since the product side (2 mol) has fewer gas moles than the reactant side (4 mol), increasing the pressure shifts the equilibrium to the right, forming more ammonia.
   4. This has nothing to do with a temperature change or with the type of bonding in ammonia; it depends only on comparing the number of gas moles on each side, so C is correct.
   • Why not A: Assumes increasing pressure always pushes an equilibrium backwards, without checking which side of the equation has fewer moles of gas; here the forward direction is actually favoured because it has fewer gas molecules.
   • Why not B: Wrongly claims pressure changes only matter for ionic substances. Le Chatelier's principle about total moles of gas applies to this reaction because both reactants and the product are gases, regardless of bonding type.
   • Why not D: Confuses a pressure change with a temperature change; increasing pressure does not, by itself, raise the temperature of the system, so this reasoning borrows the (correct) conclusion for the wrong reason.

5. Question 5Answer: B

   1. Find the relative molar mass of sodium chloride: Mr(NaCl) = Ar(Na) + Ar(Cl) = 23 + 35.5 = 58.5.
   2. The number of moles in a given mass is found using moles = mass (g) / Mr, so moles = 117 / 58.5.
   3. 117 / 58.5 = 2, since 58.5 x 2 = 117.
   4. So 117 g of sodium chloride is 2 mol, and this amount contains 2 times Avogadro's number of formula units of NaCl, since one mole of any substance always contains Avogadro's number of particles. The answer is B.
   • Why not A: Inverts the moles formula, calculating Mr divided by mass (58.5 / 117) instead of mass divided by Mr.
   • Why not C: Misplaces a decimal point in the relative molar mass, using 5.85 instead of the correct value 58.5, which multiplies the true answer by ten.
   • Why not D: Multiplies the mass by the relative molar mass instead of dividing by it, treating moles = mass x Mr rather than moles = mass / Mr.

6. Question 6Answer: A

   1. The relative molar mass of magnesium oxide is Mr(MgO) = Ar(Mg) + Ar(O) = 24 + 16 = 40.
   2. Percentage composition by mass of an element = (Ar of that element / Mr of the compound) x 100.
   3. For magnesium: (24 / 40) x 100 = 60%.
   4. So magnesium makes up 60% of the mass of magnesium oxide, option A.
   • Why not B: Calculates the percentage of oxygen in the compound (16 / 40 x 100 = 40%) rather than the percentage of magnesium that the question actually asks for.
   • Why not C: Divides the Ar of oxygen by the Ar of magnesium (16 / 24) instead of dividing the Ar of magnesium by the compound's Mr (24 / 40), then multiplies by 100.
   • Why not D: Reports the Ar value of magnesium (24) directly as though it were already a percentage, without dividing by the compound's Mr at all.

7. Question 7Answer: D

   1. Convert each mass percentage to moles by dividing by the relevant Ar: moles Fe = 70 / 56 = 1.25 mol, and moles O = 30 / 16 = 1.875 mol.
   2. Divide both mole values by the smaller one (1.25) to find the simplest ratio: Fe : O = 1.25/1.25 : 1.875/1.25 = 1 : 1.5.
   3. A ratio of 1 : 1.5 is not yet in whole numbers, so multiply both parts by 2 to clear the fraction: 2 : 3.
   4. This gives the empirical formula Fe2O3, option D. (If a relative molecular mass were also given, the same method extends to a molecular formula: divide the given Mr by the empirical formula's own mass to find how many times the empirical formula's ratio repeats.)
   • Why not A: Correctly finds the mole ratio Fe : O = 1 : 1.5, but then rounds 1.5 down to 1 instead of multiplying the whole ratio by 2 to clear the fraction and reach the smallest whole-number ratio, 2 : 3.
   • Why not B: Rounds each mole value (1.25 mol Fe and 1.875 mol O) independently to the nearest whole number (1 and 2), instead of dividing both by the smaller value to find the ratio between them.
   • Why not C: Uses the mass percentages (70 and 30) directly as the mole ratio, skipping the essential step of dividing each percentage by its element's Ar.

8. Question 8Answer: C

   1. Find the moles of each reactant: moles N2 = 28 / 28 = 1 mol, and moles H2 = 3 / 2 = 1.5 mol.
   2. The balanced equation N2 + 3H2 -> 2NH3 needs 3 mol of H2 for every 1 mol of N2. With 1 mol of N2 available, 3 mol of H2 would be needed, but only 1.5 mol of H2 is present, so hydrogen is the limiting reactant.
   3. Use the hydrogen amount to find the ammonia produced: 1.5 mol H2 x (2 mol NH3 / 3 mol H2) = 1 mol NH3.
   4. Convert to mass: Mr(NH3) = 14 + 3 = 17, so 1 mol NH3 has a mass of 1 x 17 = 17 g, option C.
   • Why not A: Assumes nitrogen is the reactant that runs out, and calculates the ammonia produced from all 1 mol of N2 reacting completely (1 mol N2 gives 2 mol NH3, or 34 g), without first checking whether there is enough hydrogen available to achieve this.
   • Why not B: Correctly identifies hydrogen as the reactant that runs out first, but then misreads the equation's mole ratio, treating NH3 : H2 as 1 : 2 (giving 1.5 / 2 = 0.75 mol NH3) instead of the correct 2 : 3 ratio from the balanced equation.
   • Why not D: Averages the two possible answers, (17 + 34) / 2 = 25.5, rather than working out which reactant actually limits the amount of product formed.

9. Question 9Answer: B

   1. Convert the given masses to moles: moles Mg = 2.4 / 24 = 0.1 mol, and moles of oxygen atoms = 1.6 / 16 = 0.1 mol.
   2. Oxygen gas is diatomic (O2), so 0.1 mol of oxygen atoms is only 0.05 mol of O2 molecules.
   3. The mole ratio of Mg to O2 is therefore 0.1 : 0.05, which simplifies to 2 : 1.
   4. Magnesium forms the Mg^2+ ion and oxygen forms the O^2- ion, which combine 1 : 1 in the product, so the balanced equation is 2Mg + O2 -> 2MgO, option B.
   • Why not A: Treats the 0.1 mol of oxygen atoms calculated from the mass data as though it were 0.1 mol of O2 molecules, forgetting that oxygen gas is diatomic; this halves the true amount of O2 needed and gives a 1 : 1 ratio of Mg to O2 instead of the correct 2 : 1.
   • Why not C: Correctly works out the 2 : 1 mole ratio of Mg to O2, but then writes the product as the peroxide-style formula MgO2, forgetting that Mg^2+ and O^2- ions combine in a 1 : 1 ratio to give MgO, not MgO2.
   • Why not D: Miscounts the ionic ratio in the product formula, writing Mg2O instead of the correct 1 : 1 pairing of Mg^2+ and O^2- ions in magnesium oxide, MgO.

10. Question 10Answer: A

    1. Find the relative molecular mass of carbon dioxide: Mr(CO2) = Ar(C) + 2 x Ar(O) = 12 + (2 x 16) = 44.
    2. Convert the given mass to moles: moles = mass / Mr = 4.4 / 44 = 0.1 mol.
    3. At room temperature and pressure, one mole of any gas occupies 24 dm3, so the volume of 0.1 mol is 0.1 x 24.
    4. 0.1 x 24 = 2.4, so the volume occupied is 2.4 dm3, option A.
    • Why not B: Forgets to convert the given mass into moles first, treating the 4.4 g sample as though it were already exactly one mole and reporting the molar volume directly.
    • Why not C: Correctly finds 0.1 mol but slips a decimal place when multiplying by the molar volume, computing 0.1 x 2.4 instead of 0.1 x 24.
    • Why not D: Confuses the relative molecular mass, Mr = 44, with the volume being asked for, and simply restates that number in dm3.

11. Question 11Answer: D

    1. Find the moles of HCl used: moles = concentration x volume (in dm3) = 0.50 x (20.0 / 1000) = 0.50 x 0.0200 = 0.01 mol.
    2. The equation NaOH + HCl -> NaCl + H2O shows a 1 : 1 mole ratio, so 0.01 mol of HCl reacts with exactly 0.01 mol of NaOH.
    3. The volume of the sodium hydroxide solution is 25.0 cm3, which is 0.0250 dm3.
    4. Concentration of NaOH = moles / volume (dm3) = 0.01 / 0.0250 = 0.40 mol dm-3, option D.
    • Why not A: Inverts the volume ratio used to scale the concentration, calculating 0.50 x (25.0 / 20.0) instead of 0.50 x (20.0 / 25.0).
    • Why not B: Reports the number of moles of acid used in the titration, 0.01 mol, directly as though it were the concentration, forgetting to divide by the volume of the sodium hydroxide solution.
    • Why not C: Assumes the two solutions must have equal concentrations simply because they exactly neutralise each other, ignoring that their volumes (20.0 cm3 and 25.0 cm3) are different.

12. Question 12Answer: C

    1. Percentage yield = (actual yield / predicted yield) x 100.
    2. Here, actual yield = 10.0 g and predicted yield = 12.5 g.
    3. 10.0 / 12.5 = 0.8, since 12.5 x 0.8 = 10.0.
    4. 0.8 x 100 = 80%, so the percentage yield is 80%, option C.
    • Why not A: Inverts the percentage yield formula, calculating predicted yield divided by actual yield (12.5 / 10.0 x 100) instead of actual divided by predicted.
    • Why not B: Calculates the percentage of the product that was lost (the 2.5 g shortfall as a fraction of the predicted mass: 2.5 / 12.5 x 100 = 20%) instead of the percentage that was actually obtained.
    • Why not D: Uses only the difference between the predicted and actual masses, 2.5 g, as though it were already the percentage, without dividing by the predicted mass at all.

13. Question 13Answer: B

    1. Reduction can be defined either as the loss of oxygen or as the gain of electrons by a species.
    2. In Fe2O3 + 3CO -> 2Fe + 3CO2, the iron starts combined with oxygen (in Fe2O3) and ends up as pure metal (Fe), so it has lost its oxygen: this is reduction.
    3. In electron terms, each Fe^3+ ion in Fe2O3 gains 3 electrons to become a neutral Fe atom, which also confirms that reduction is happening to the iron.
    4. Meanwhile CO gains oxygen to become CO2, which is oxidation, not reduction, so the reduced species is Fe2O3 (or the iron within it), option B.
    • Why not A: Overgeneralises 'gaining something means reduction' to mean gaining any atom at all. In fact CO gaining oxygen atoms to become CO2 is the definition of oxidation, not reduction; CO is oxidised here, not reduced.
    • Why not C: Confuses the everyday meaning of 'reduced' (made smaller, less massive) with the chemical definition, which is about losing oxygen or gaining electrons, not about an overall decrease in mass.
    • Why not D: Invents a definition of reduction based on molecular complexity, which is not a real chemical definition; reduction is defined by the gain of electrons or the loss of oxygen, not by how complicated a molecule's structure is.

14. Question 14Answer: A

    1. Sulfate is a polyatomic ion with a fixed overall charge of -2, written SO4^2-.
    2. Since CuSO4 is a neutral compound, the oxidation state of copper, x, must satisfy x + (-2) = 0, so x = +2.
    3. Zinc starts as the element (oxidation state 0) and ends up as Zn^2+ in ZnSO4 (oxidation state +2): its oxidation state increases, so zinc is oxidised.
    4. Copper starts as Cu^2+ in CuSO4 (oxidation state +2) and ends up as the element Cu (oxidation state 0): its oxidation state decreases, so copper is reduced. Because both an oxidation and a reduction happen together, this is a redox reaction, option A.
    • Why not B: Copies the charge of the polyatomic sulfate ion (-2) straight onto copper, instead of solving for copper's oxidation state so that the whole neutral compound balances: x + (-2) = 0 gives x = +2, not -2.
    • Why not C: Correctly finds copper's oxidation state but wrongly restricts the classification to zinc's change alone; a redox reaction requires both an oxidation and a paired reduction, and copper's change from +2 to 0 is the reduction half that this option misses.
    • Why not D: Ignores that the sulfate ion, SO4^2-, has a fixed overall charge of -2 as a single polyatomic unit, and instead wrongly adds up four separate oxygen charges of -2 each (a total of -8) as if none of that charge were already accounted for within the sulfate ion.

15. Question 15Answer: C

    1. Disproportionation is when atoms of the same element are simultaneously oxidised and reduced within one reaction.
    2. Here, chlorine starts at oxidation state 0 in Cl2, then ends up at -1 in NaCl (a decrease, so reduction) and at +1 in NaOCl (an increase, so oxidation).
    3. Because both changes happen to chlorine atoms within the same reaction, chlorine (Cl2) is acting as both agents at once: the chlorine atoms being reduced act as the oxidising agent towards the other chlorine atoms, while those being oxidised act as the reducing agent, and since it is all the same element, Cl2 is described as both.
    4. Sodium, oxygen and hydrogen keep the same oxidation states throughout (Na +1, O -2, H +1), so sodium hydroxide plays no redox role; the correct description is that chlorine (Cl2) is both agents, option C.
    • Why not A: Correctly names the phenomenon as disproportionation, but assigns the 'both agent' role to sodium hydroxide, a species whose own elements (Na +1, O -2, H +1) do not change oxidation state at all in this reaction.
    • Why not B: Confuses 'two different products formed' with 'two different elements being redox-active'. Disproportionation is defined by one element being simultaneously oxidised and reduced, which is exactly what happens to chlorine here, regardless of how many product species end up containing it.
    • Why not D: Invents an oxidation state change for hydrogen where none actually occurs: hydrogen is +1 in NaOH before the reaction and remains +1 in H2O afterwards, so there is no change to point to.