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Admissions tests / ESAT / Chemistry / Quantitative chemistry and redox

Test standard. 15 questions, 15 marks, about 22 minutes.

ESAT Chemistry: Quantitative chemistry and redox, set 2

Chemical equations, relative molar mass, moles, concentration, yields, limiting reagents, empirical formulae, and oxidation and reduction.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  • Use the Ar values given in each question.
  1. 11 mark

    In a sealed container, 12.0 g of carbon is burned completely in an excess of oxygen gas to form carbon dioxide. Given that the total mass of carbon dioxide produced is 44.0 g, what mass of oxygen combined with the carbon?

    1. A 44.0 g
    2. B 56.0 g
    3. C 32.0 g
    4. D 20.0 g
  2. 21 mark

    Iron(III) ions form Fe^3+ and nitrate ions form NO3^-. What is the correct formula, including state symbol, for solid iron(III) nitrate?

    1. A Fe(NO3)3 (s)
    2. B FeNO3 (s)
    3. C Fe3(NO3) (s)
    4. D Fe(NO3)3 (aq)
  3. 31 mark

    Iron(II) ions are oxidised to iron(III) ions during a reaction. Which half-equation correctly represents this change?

    1. A Fe^2+ + e- -> Fe^3+
    2. B Fe^3+ + e- -> Fe^2+
    3. C Fe^2+ -> Fe^3+ + 2e-
    4. D Fe^2+ -> Fe^3+ + e-
  4. 41 mark

    Aluminium reacts with dilute sulfuric acid: Al + H2SO4 -> Al2(SO4)3 + H2. When this equation is balanced using the smallest possible whole-number coefficients, what is the coefficient of H2SO4?

    1. A 1
    2. B 3
    3. C 2
    4. D 6
  5. 51 mark

    The reaction 2SO2(g) + O2(g) <=> 2SO3(g) is exothermic in the forward direction. What is the effect of increasing the temperature, at constant pressure, on the position of equilibrium and the yield of SO3?

    1. A The equilibrium shifts left, decreasing the yield of SO3, because increasing temperature favours the endothermic direction of a reversible reaction, and here that is the reverse reaction.
    2. B The equilibrium shifts right, increasing the yield of SO3, because raising the temperature speeds up the forward reaction more than the reverse reaction, and a faster forward reaction must produce more product overall.
    3. C The equilibrium position is not affected by the temperature change, because temperature, unlike pressure or concentration, only affects how quickly equilibrium is reached, not where the equilibrium lies.
    4. D The equilibrium shifts right, increasing the yield of SO3, because raising the temperature of a gas mixture also raises its pressure, and increasing pressure favours the side of the equation with fewer moles of gas.
  6. 61 mark

    Ammonium nitrate, NH4NO3, is used as a fertiliser. Use Ar: N = 14, H = 1, O = 16. What is the percentage by mass of nitrogen in ammonium nitrate?

    1. A 17.5%
    2. B 14%
    3. C 35%
    4. D 60%
  7. 71 mark

    2.7 g of aluminium reacts completely with oxygen gas to form 5.1 g of aluminium oxide, with no aluminium or oxygen left over. Use Ar: Al = 27, O = 16. What is the empirical formula of aluminium oxide, based on this data?

    1. A AlO
    2. B Al3O2
    3. C AlO2
    4. D Al2O3
  8. 81 mark

    Methane burns in oxygen: CH4 + 2O2 -> CO2 + 2H2O. 16 g of methane is mixed with 48 g of oxygen and ignited. Use Ar: C = 12, H = 1, O = 16. What is the maximum mass of carbon dioxide that can be formed?

    1. A 9 g
    2. B 33 g
    3. C 44 g
    4. D 66 g
  9. 91 mark

    Hydrogen gas reacts with chlorine gas to form hydrogen chloride gas. In an experiment, 40 cm3 of hydrogen reacts exactly with 40 cm3 of chlorine, with all volumes measured at the same temperature and pressure, so the volumes are directly proportional to the number of moles present. Which balanced equation is consistent with this data?

    1. A H2 + Cl2 -> 2HCl
    2. B 2H2 + Cl2 -> 2HCl
    3. C H2 + Cl2 -> HCl
    4. D H2 + 2Cl2 -> 2HCl
  10. 101 mark

    A gas syringe contains 6 dm3 of methane gas, CH4, at room temperature and pressure, where one mole of any gas occupies 24 dm3 under these conditions. Use Ar: C = 12, H = 1. What is the mass of methane in the syringe?

    1. A 3 g
    2. B 96 g
    3. C 64 g
    4. D 4 g
  11. 111 mark

    A solution of sodium hydroxide, NaOH, has a concentration of 2.0 mol dm-3. Use Ar: Na = 23, O = 16, H = 1. What is the concentration of this solution in g dm-3?

    1. A 20 g dm-3
    2. B 8 g dm-3
    3. C 80 g dm-3
    4. D 42 g dm-3
  12. 121 mark

    In a titration, 20.0 cm3 of sulfuric acid exactly neutralises 50.0 cm3 of 0.20 mol dm-3 sodium hydroxide solution: 2NaOH + H2SO4 -> Na2SO4 + 2H2O. What is the concentration of the sulfuric acid, in mol dm-3?

    1. A 0.005 mol dm-3
    2. B 0.25 mol dm-3
    3. C 1.0 mol dm-3
    4. D 0.50 mol dm-3
  13. 131 mark

    Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide: CaCO3 -> CaO + CO2. Use Ar: Ca = 40, C = 12, O = 16. If calcium oxide is the desired product, what is the percentage atom economy of this reaction?

    1. A 44%
    2. B 56%
    3. C 100%
    4. D 40%
  14. 141 mark

    In potassium manganate(VII), KMnO4, potassium has an oxidation state of +1 and oxygen has an oxidation state of -2, and the overall compound is neutral. What is the oxidation state of manganese in this compound?

    1. A +7
    2. B -7
    3. C +8
    4. D +5
  15. 151 mark

    Chlorine gas displaces bromine from potassium bromide solution: Cl2 + 2KBr -> 2KCl + Br2. Which statement correctly identifies the oxidising agent and the reducing agent in this reaction?

    1. A Chlorine is oxidised and is therefore called the oxidising agent, while bromide is reduced and is therefore called the reducing agent.
    2. B Chlorine is the reducing agent, because it is the reactant that is added to the solution to start the reaction.
    3. C Chlorine (Cl2) is the oxidising agent, since it is reduced (each chlorine atom gains an electron to form a chloride ion); potassium bromide is the reducing agent, since bromide is oxidised (each bromide ion loses an electron to form bromine).
    4. D Potassium bromide is the oxidising agent, because bromide ions are the ones that end up leaving the compound during the reaction.

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: C

    1. The container is sealed, so no matter can enter or leave: the law of conservation of mass says the total mass of the reactants (carbon plus the oxygen that reacts with it) must equal the total mass of the product formed.
    2. Total reactant mass = total product mass, so mass of carbon + mass of oxygen that reacted = mass of carbon dioxide produced.
    3. Rearranging: mass of oxygen = mass of carbon dioxide - mass of carbon = 44.0 - 12.0 = 32.0 g.
    4. So 32.0 g of oxygen combined with the carbon, option C. (This also matches the stoichiometry of C + O2 -> CO2: 12.0 g of carbon is 1 mol, which needs exactly 1 mol of O2, 32.0 g, to give 1 mol of CO2, 44.0 g.)
    • Why not A: Uses the total mass of carbon dioxide produced directly as though it were the mass of oxygen that reacted, forgetting to subtract the mass of the carbon that is already accounted for inside that 44.0 g.
    • Why not B: Adds the two given masses together instead of subtracting them, treating conservation of mass as though the product mass should exceed the sum of the reactants rather than equal it.
    • Why not D: Subtracts the mass of carbon twice, 44.0 - 12.0 - 12.0, as though the carbon's mass needed to be removed from the product a second time after it had already been accounted for once.

  2. Question 2Answer: A

    1. Iron(III) forms the Fe^3+ ion, and nitrate is the polyatomic ion NO3^-, carrying a single negative charge.
    2. To balance the charges, the size of one ion's charge becomes the number of the other ion needed: one Fe^3+ ion (charge +3) requires three NO3^- ions (total charge -3) to give a neutral compound.
    3. This gives the formula Fe(NO3)3, with brackets around the nitrate group to show that the whole NO3 unit is repeated three times.
    4. The question asks for the solid form of the compound, so the state symbol is (s), giving Fe(NO3)3 (s), option A.
    • Why not B: Writes a 1:1 ratio of iron to nitrate ions, ignoring that three NO3^- ions (each carrying a single negative charge) are needed to balance the +3 charge on one Fe^3+ ion.
    • Why not C: Crosses the ions' charges over the wrong way, placing the 3 from iron's own 3+ charge as iron's own subscript instead of using it as the number of nitrate ions needed to balance that charge.
    • Why not D: Gets the correct formula unit but the wrong state symbol; the question specifies the solid compound, so it should be labelled (s), not (aq) for a solution.

  3. Question 3Answer: D

    1. Oxidation is the loss of electrons, so the oxidation of Fe^2+ to Fe^3+ must show an electron appearing as a product, on the right-hand side of the half-equation.
    2. Fe^2+ has charge +2; forming Fe^3+ (charge +3) means the iron ion has become more positively charged, which happens by losing one negatively charged electron.
    3. Check the charge balance: left-hand side is +2; right-hand side is Fe^3+ (+3) plus e- (-1), which is +3 - 1 = +2. The two sides match.
    4. So the correct half-equation is Fe^2+ -> Fe^3+ + e-, option D.
    • Why not A: Places the electron on the reactant side, as though iron were gaining an electron during oxidation, while still showing the charge increasing to 3+. Checking charge shows this cannot balance: the left side is (+2) + (-1) = +1, but the right side is +3. Oxidation is a loss of electrons, so the electron lost must appear as a product, on the right-hand side.
    • Why not B: Writes the reduction half-equation instead of the oxidation half-equation the question asks for, reversing which species is losing electrons and which is gaining them.
    • Why not C: Assumes iron always loses two electrons when it is oxidised, as if every change in oxidation state involved two electrons; but here the oxidation state only changes by one unit, from +2 to +3, so only one electron is lost. Checking charge confirms the error: (+2) does not equal (+3) + (-2) = +1.

  4. Question 4Answer: B

    1. Balance aluminium first: Al2(SO4)3 contains 2 aluminium atoms, so 2 Al is needed on the left.
    2. Balance the sulfate groups next: Al2(SO4)3 contains 3 separate SO4 groups, and each H2SO4 supplies one SO4 group, so 3 H2SO4 are needed on the left.
    3. Check hydrogen: 3 H2SO4 supplies 3 x 2 = 6 hydrogen atoms, so 3 H2 are needed on the right (3 x 2 = 6 hydrogen atoms). Check oxygen: 3 H2SO4 supplies 3 x 4 = 12 oxygen atoms, matching the 3 x 4 = 12 oxygen atoms in Al2(SO4)3.
    4. The balanced equation is 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2, so the coefficient of H2SO4 is 3, option B.
    • Why not A: Leaves every coefficient at 1, as the equation is first written, without checking that the atoms of every element actually balance on both sides.
    • Why not C: Matches the coefficient of H2SO4 to the coefficient of aluminium (2), assuming a 1:1 ratio between aluminium atoms and formula units of sulfuric acid instead of working out the ratio the sulfate groups actually require.
    • Why not D: Doubles the number of sulfate groups needed, treating the subscript 3 in Al2(SO4)3 as though it also had to be multiplied by the 2 aluminium atoms, giving 3 x 2 = 6 instead of just 3.

  5. Question 5Answer: A

    1. If a reversible reaction is exothermic in the forward direction, it must be endothermic in the reverse direction (the two directions always have opposite sign of enthalpy change).
    2. Le Chatelier's principle says that increasing temperature shifts an equilibrium in the endothermic direction, since that direction absorbs some of the added energy.
    3. Here, the forward reaction is exothermic, so the reverse reaction (2SO3 -> 2SO2 + O2) is the endothermic direction.
    4. Increasing temperature therefore shifts the equilibrium left, towards SO2 and O2, decreasing the yield of SO3, option A. This has nothing to do with the number of gas moles on each side, since pressure is held constant.
    • Why not B: Confuses rate with equilibrium position: raising temperature speeds up both the forward and reverse reactions, and it is specifically which direction is endothermic, not which one 'speeds up more', that decides which way the equilibrium position shifts.
    • Why not C: Wrongly excludes temperature from the factors that shift equilibrium position. Unlike a catalyst, which only changes how fast equilibrium is reached, temperature is the one factor that changes the value of the equilibrium constant itself and so does move the position of equilibrium.
    • Why not D: Confuses a change in temperature with a change in pressure. The question states that pressure is held constant, so reasoning about pressure changing (even though it does correctly identify that fewer gas moles sit on the product side) smuggles in a variable that was explicitly fixed.

  6. Question 6Answer: C

    1. NH4NO3 contains 2 nitrogen atoms (one in NH4, one in NO3), 4 hydrogen atoms and 3 oxygen atoms.
    2. Relative formula mass: Mr(NH4NO3) = (2 x 14) + (4 x 1) + (3 x 16) = 28 + 4 + 48 = 80.
    3. Percentage composition by mass of an element = (total Ar of that element in the formula / Mr of the compound) x 100.
    4. For nitrogen: (28 / 80) x 100 = 35%, option C.
    • Why not A: Uses only one nitrogen atom, 14 / 80 x 100 = 17.5%, forgetting that ammonium nitrate contains two nitrogen atoms: one in the ammonium ion and one in the nitrate ion.
    • Why not B: States the relative atomic mass of nitrogen directly as though it were already the percentage, without dividing by the compound's relative formula mass at all.
    • Why not D: Calculates the percentage of oxygen instead of the percentage of nitrogen that the question asks for: 48 / 80 x 100 = 60%.

  7. Question 7Answer: D

    1. By conservation of mass, the mass of oxygen that reacted = 5.1 - 2.7 = 2.4 g.
    2. Convert each mass to moles: moles Al = 2.7 / 27 = 0.1 mol; moles O = 2.4 / 16 = 0.15 mol.
    3. Divide both mole values by the smaller one (0.1) to find the simplest ratio: Al : O = 1 : 1.5. This is not yet whole numbers, so multiply both parts by 2 to clear the fraction: 2 : 3.
    4. This gives the empirical formula Al2O3, option D, which is indeed the real formula of aluminium oxide.
    • Why not A: Correctly finds the mole ratio Al : O = 1 : 1.5 (after dividing both mole values by the smaller one), but then rounds 1.5 down to 1 instead of multiplying the whole ratio by 2 to clear the fraction.
    • Why not B: Correctly simplifies the ratio to the whole numbers 2 : 3, but then writes the two numbers the wrong way round, swapping which one belongs to aluminium and which to oxygen.
    • Why not C: Finds the same ratio 1 : 1.5 but rounds 1.5 up to 2 instead of multiplying through by 2, giving a ratio of 1 : 2 rather than the correct 2 : 3.

  8. Question 8Answer: B

    1. Find the moles of each reactant: moles CH4 = 16 / 16 = 1 mol (Mr of CH4 = 12 + 4 = 16); moles O2 = 48 / 32 = 1.5 mol.
    2. The equation CH4 + 2O2 -> CO2 + 2H2O needs 2 mol of O2 for every 1 mol of CH4. With 1 mol of CH4 available, 2 mol of O2 would be needed, but only 1.5 mol of O2 is present, so oxygen is the limiting reactant.
    3. Use the oxygen amount to find the carbon dioxide produced: 1.5 mol O2 x (1 mol CO2 / 2 mol O2) = 0.75 mol CO2.
    4. Convert to mass: Mr(CO2) = 12 + (2 x 16) = 44, so 0.75 mol CO2 has a mass of 0.75 x 44 = 33 g, option B.
    • Why not A: Uses the correct limiting reactant and the correct 0.75 mol of product, but converts moles to mass using the relative atomic mass of carbon alone (12) instead of the relative formula mass of carbon dioxide (44): 0.75 x 12 = 9 g.
    • Why not C: Assumes methane is the reactant that runs out, and calculates the carbon dioxide produced from all 1 mol of methane reacting completely, without first checking whether enough oxygen is actually available.
    • Why not D: Uses the wrong mole ratio between oxygen and carbon dioxide, treating them as 1 : 1 instead of 2 : 1, so takes all 1.5 mol of available oxygen directly as 1.5 mol of carbon dioxide.

  9. Question 9Answer: A

    1. Equal volumes of gases at the same temperature and pressure contain equal numbers of moles, so the 40 cm3 : 40 cm3 ratio of hydrogen to chlorine is also the mole ratio, 1 : 1.
    2. A balanced equation with a 1 : 1 ratio of H2 to Cl2 that also balances the atoms is H2 + Cl2 -> 2HCl: 2 hydrogen atoms and 2 chlorine atoms on the left match the 2 hydrogen and 2 chlorine atoms in 2HCl on the right.
    3. None of the other options both match the 1:1 volume ratio and balance every atom, as their individual rationales show.
    4. So the equation consistent with the data is H2 + Cl2 -> 2HCl, option A.
    • Why not B: Does not balance: there are 4 hydrogen atoms on the left but only 2 on the right, since an extra H2 has been added that the 1:1 volume data gives no reason to include.
    • Why not C: Does not balance: both hydrogen and chlorine appear as 2 atoms on the left but only as 1 atom's worth on the right, forgetting that the HCl coefficient must be doubled to balance both atoms at once.
    • Why not D: Does not balance: there are 4 chlorine atoms on the left but only 2 on the right, misreading the equal 40:40 volume data as though chlorine were present in double the amount of hydrogen.

  10. Question 10Answer: D

    1. At room temperature and pressure, one mole of any gas occupies 24 dm3, so the number of moles of methane present is moles = volume / 24 = 6 / 24 = 0.25 mol.
    2. The relative formula mass of methane is Mr(CH4) = 12 + (4 x 1) = 16.
    3. Mass = moles x Mr = 0.25 x 16.
    4. 0.25 x 16 = 4, so the mass of methane in the syringe is 4 g, option D.
    • Why not A: Correctly finds 0.25 mol but converts to mass using the relative atomic mass of carbon alone (12) instead of the relative formula mass of methane (16): 0.25 x 12 = 3 g.
    • Why not B: Uses the volume figure directly as though it were already the number of moles, multiplying 6 by the relative formula mass without first dividing by the molar gas volume: 6 x 16 = 96 g.
    • Why not C: Inverts the molar-volume division, calculating moles as 24 / 6 = 4 instead of 6 / 24 = 0.25, then multiplies by the relative formula mass: 4 x 16 = 64 g.

  11. Question 11Answer: C

    1. The relative formula mass of sodium hydroxide is Mr(NaOH) = 23 + 16 + 1 = 40.
    2. Concentration in g dm-3 = concentration in mol dm-3 x Mr, since each mole of solute has a mass of Mr grams.
    3. Concentration in g dm-3 = 2.0 x 40.
    4. 2.0 x 40 = 80, so the concentration is 80 g dm-3, option C.
    • Why not A: Divides the relative formula mass by the concentration instead of multiplying them together: 40 / 2.0 = 20 g dm-3.
    • Why not B: Multiplies by only a tenth of the correct relative formula mass, as though Mr(NaOH) were 4 instead of 40: 2.0 x 4 = 8 g dm-3.
    • Why not D: Adds the concentration value to the relative formula mass instead of multiplying them together: 40 + 2 = 42 g dm-3.

  12. Question 12Answer: B

    1. Find the moles of NaOH used: moles = concentration x volume (in dm3) = 0.20 x (50.0 / 1000) = 0.20 x 0.0500 = 0.01 mol.
    2. The equation 2NaOH + H2SO4 -> Na2SO4 + 2H2O shows a 2:1 mole ratio of NaOH to H2SO4, so the moles of H2SO4 are half the moles of NaOH: 0.01 / 2 = 0.005 mol.
    3. The volume of sulfuric acid is 20.0 cm3, which is 0.0200 dm3.
    4. Concentration of H2SO4 = moles / volume (dm3) = 0.005 / 0.0200 = 0.25 mol dm-3, option B.
    • Why not A: Reports the number of moles of sulfuric acid used, 0.005 mol, directly as though it were already the concentration, forgetting to divide by the volume of acid used.
    • Why not C: Inverts the mole ratio from the balanced equation, doubling instead of halving the moles of NaOH to find the moles of H2SO4 (perhaps from over-applying the idea that sulfuric acid is diprotic), giving 0.01 x 2 = 0.02 mol and so 0.02 / 0.020 = 1.0 mol dm-3.
    • Why not D: Ignores the balanced equation's 2:1 mole ratio and treats moles of H2SO4 as equal to moles of NaOH (0.01 mol), giving 0.01 / 0.020 = 0.50 mol dm-3.

  13. Question 13Answer: B

    1. Percentage atom economy = (relative formula mass of the desired product / sum of the relative formula masses of all the products) x 100.
    2. Relative formula mass of CaCO3 = 40 + 12 + (3 x 16) = 40 + 12 + 48 = 100.
    3. Relative formula mass of CaO (the desired product) = 40 + 16 = 56. Relative formula mass of CO2 (the by-product) = 12 + (2 x 16) = 44. Check: 56 + 44 = 100, which correctly equals the relative formula mass of the single reactant, CaCO3, confirming conservation of mass.
    4. Atom economy = (56 / 100) x 100 = 56%, option B.
    • Why not A: Uses the relative formula mass of carbon dioxide, the by-product, as the numerator instead of the relative formula mass of calcium oxide, the stated desired product: 44 / 100 x 100 = 44%.
    • Why not C: Assumes atom economy must always be 100% because no atoms are created or destroyed in the reaction. Conservation of mass guarantees the total mass of all the products together equals the mass of the reactant, but atom economy only counts the mass of the DESIRED product, not every product formed.
    • Why not D: Calculates the percentage by mass of calcium within calcium carbonate itself, 40 / 100 x 100 = 40%, a percentage composition calculation, rather than the atom economy of the reaction, which compares the desired product's mass to the total mass of all the products formed.

  14. Question 14Answer: A

    1. The oxidation states in a neutral compound must add up to zero.
    2. KMnO4 contains 1 potassium atom (+1), 1 manganese atom (Mn, unknown) and 4 oxygen atoms (-2 each, totalling -8).
    3. So 1 + Mn + (-8) = 0, which rearranges to Mn = 8 - 1 = 7.
    4. The oxidation state of manganese in KMnO4 is +7, option A.
    • Why not B: Adds the other oxidation states together directly as manganese's own value, 1 + (-8) = -7, instead of solving so that the whole sum equals zero, which requires manganese to be the positive value that cancels this out.
    • Why not C: Forgets to include potassium's +1 contribution, solving only 4 x (-2) + Mn = 0, which gives Mn = +8.
    • Why not D: Miscounts the number of oxygen atoms in the formula as three instead of four, solving 1 + Mn + 3 x (-2) = 0, which gives Mn = +5.

  15. Question 15Answer: C

    1. Chlorine starts at oxidation state 0 in Cl2 and ends at -1 in KCl: this is a decrease in oxidation state, so chlorine is reduced.
    2. Bromine starts at oxidation state -1 in KBr and ends at 0 in Br2: this is an increase in oxidation state, so bromide is oxidised.
    3. The species that is reduced is called the oxidising agent, because it causes oxidation in the other species by accepting electrons from it; the species that is oxidised is called the reducing agent, because it causes reduction in the other species by donating electrons to it.
    4. So chlorine (Cl2), which is reduced, is the oxidising agent, and potassium bromide (the bromide ion), which is oxidised, is the reducing agent, option C.
    • Why not A: The classic mix-up: it matches each agent's name to what happens to the substance itself, when the name actually describes what the substance does to the OTHER reactant. The species that is oxidised is called the reducing agent (it reduces something else by giving up electrons), and the species that is reduced is called the oxidising agent (it oxidises something else by taking electrons) - the opposite pairing to this option.
    • Why not B: Names the agent based on which substance was added or caused the reaction to start, rather than on which species gains electrons. Chlorine gains electrons here (it is reduced), which is what makes it the oxidising agent, regardless of which reactant was introduced first.
    • Why not D: Names the agent based on which ion 'leaves' the compound rather than on the direction of electron transfer. Bromide is in fact oxidised in this reaction, which makes potassium bromide the reducing agent, not the oxidising agent.

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