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Admissions tests / ESAT / Chemistry / Quantitative chemistry and redox

Stretch. 15 questions, 15 marks, about 31 minutes.

ESAT Chemistry: Quantitative chemistry and redox, set 4

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.
  • This is a stretch set: most questions combine two specification points, and the fastest route is rarely the first one you see.
  1. 11 mark

    Copper(II) oxide is reduced by hydrogen: CuO(s) + H2(g) -> Cu(s) + H2O(g).

    40 g of copper(II) oxide is heated in a stream containing 0.6 g of hydrogen, and the reaction is allowed to go to completion. The copper recovered weighs less than the maximum possible because some is lost on the apparatus; the yield for this preparation is 75%.

    Using Ar: Cu = 64, O = 16, H = 1, what mass of copper is recovered?

    1. A 14.4 g
    2. B 19.2 g
    3. C 24.0 g
    4. D 32.0 g
  2. 21 mark

    A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, and its relative molecular mass is 180.

    Using Ar: C = 12, H = 1, O = 16, what is its molecular formula?

    1. A CH2O
    2. B C2H4O2
    3. C C6H12O6
    4. D C3H6O3
  3. 31 mark

    Sulfuric acid neutralises sodium hydroxide: H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l).

    In a titration, 25.0 cm3 of sodium hydroxide solution is exactly neutralised by 20.0 cm3 of 0.10 mol dm-3 sulfuric acid.

    Using Ar: Na = 23, O = 16, H = 1, what is the concentration of the sodium hydroxide solution in g dm-3?

    1. A 3.2 g dm-3
    2. B 6.4 g dm-3
    3. C 12.8 g dm-3
    4. D 0.16 g dm-3
  4. 41 mark

    In a disproportionation reaction, a single element is both oxidised and reduced.

    Which one of the following reactions is a disproportionation?

    1. A 2Cu+(aq) -> Cu2+(aq) + Cu(s)
    2. B Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)
    3. C 2H2(g) + O2(g) -> 2H2O(l)
    4. D HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)
  5. 51 mark

    Ammonia is oxidised over a catalyst to make nitrogen monoxide.

    40 cm3 of ammonia reacts exactly with 50 cm3 of oxygen to produce 40 cm3 of nitrogen monoxide, with water condensed out. All gas volumes are measured at the same temperature and pressure, so they are in the same ratio as the numbers of moles.

    Which balanced equation is consistent with this data?

    1. A 2NH3(g) + 2O2(g) -> 2NO(g) + 3H2O(l)
    2. B 4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(l)
    3. C NH3(g) + O2(g) -> NO(g) + H2O(l)
    4. D 4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(l)
  6. 61 mark

    L represents Avogadro's number, 6.02 x 10^23.

    Using Ar: N = 14, H = 1, how many atoms in total are present in 3.4 g of ammonia, NH3?

    1. A 0.2L
    2. B 0.8L
    3. C 0.6L
    4. D 3.4L
  7. 71 mark

    At 25 degrees C a saturated solution of sodium chloride contains 35.1 g of dissolved sodium chloride in every 100 cm3 of solution.

    Using Ar: Na = 23, Cl = 35.5, what is the concentration of this saturated solution in mol dm-3?

    1. A 0.6 mol dm-3
    2. B 3.51 mol dm-3
    3. C 0.06 mol dm-3
    4. D 6.0 mol dm-3
  8. 81 mark

    A metal M forms an oxide with the formula M2O3, and that oxide is 70% M by mass.

    Using Ar: O = 16, what is the relative atomic mass of M?

    1. A 27
    2. B 48
    3. C 56
    4. D 112
  9. 91 mark

    Hydrogen iodide decomposes in a closed container and reaches equilibrium: 2HI(g) <=> H2(g) + I2(g). The forward reaction is endothermic.

    The overall pressure of the system is increased at constant temperature. What happens to the position of equilibrium?

    1. A It does not move, because there are equal numbers of moles of gas on each side of the equation.
    2. B It moves to the right, because the forward reaction produces two separate substances from one.
    3. C It moves to the left, because a higher pressure always favours the side with fewer molecules.
    4. D It moves to the left, because increasing the pressure has the same effect as lowering the temperature.
  10. 101 mark

    Iron(III) oxide is reduced by carbon in a blast furnace: 2Fe2O3(s) + 3C(s) -> 4Fe(l) + 3CO2(g).

    By assigning oxidation states to every element, which species acts as the reducing agent, and what happens to it?

    1. A Fe2O3, which is oxidised from 0 to +3.
    2. B Fe2O3, which is reduced from +3 to 0.
    3. C CO2, which is oxidised from 0 to +4.
    4. D C, which is oxidised from 0 to +4.
  11. 111 mark

    Hydrated copper(II) sulfate has the formula CuSO4.5H2O. On heating to constant mass it loses all of its water of crystallisation, leaving anhydrous copper(II) sulfate, CuSO4.

    Using Ar: Cu = 64, S = 32, O = 16, H = 1, what mass of anhydrous copper(II) sulfate is left when 25 g of the hydrated salt is heated to constant mass?

    1. A 9.0 g
    2. B 16.0 g
    3. C 20.0 g
    4. D 25.0 g
  12. 121 mark

    Limestone is heated in a kiln and the calcium carbonate in it decomposes completely: CaCO3(s) -> CaO(s) + CO2(g).

    A 12.5 g sample of limestone is 80% calcium carbonate by mass; the rest is inert and does not decompose.

    Using Ar: Ca = 40, C = 12, O = 16, and taking the molar volume of a gas as 24 dm3 at room temperature and pressure, what volume of carbon dioxide is produced?

    1. A 1.92 dm3
    2. B 3.00 dm3
    3. C 2.40 dm3
    4. D 0.10 dm3
  13. 131 mark

    Magnesium burns in nitrogen to form a single compound, magnesium nitride, which contains only magnesium and nitrogen.

    0.72 g of magnesium reacts completely with nitrogen to give 1.00 g of magnesium nitride, with no magnesium left over.

    Using Ar: Mg = 24, N = 14, what is the formula of magnesium nitride?

    1. A MgN
    2. B Mg2N3
    3. C MgN2
    4. D Mg3N2
  14. 141 mark

    A metal hydroxide has the formula M(OH)2 and reacts with hydrochloric acid: M(OH)2(aq) + 2HCl(aq) -> MCl2(aq) + 2H2O(l).

    A solution of this hydroxide contains 2.32 g dm-3. In a titration, 25.0 cm3 of it is exactly neutralised by 20.0 cm3 of 0.10 mol dm-3 hydrochloric acid.

    Using Ar: O = 16, H = 1, what is the relative atomic mass of M?

    1. A 24
    2. B 40
    3. C 58
    4. D 64
  15. 151 mark

    Methanol is manufactured from an equilibrium in a closed system: CO(g) + 2H2(g) <=> CH3OH(g). The forward reaction is exothermic.

    Which change would increase the equilibrium yield of methanol?

    1. A Adding a catalyst, which speeds up the forward reaction more than the reverse one.
    2. B Raising the temperature, which supplies the energy the forward reaction needs.
    3. C Raising the overall pressure, at constant temperature.
    4. D Removing some carbon monoxide from the mixture, at constant temperature.

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: A

    1. Convert both reactants to moles before comparing them, because mass alone says nothing about which runs out first. Mr(CuO) = 64 + 16 = 80, so 40 g is 40/80 = 0.5 mol. Mr(H2) = 2, so 0.6 g is 0.6/2 = 0.3 mol.
    2. The equation is 1:1 in CuO and H2, so 0.5 mol of oxide would need 0.5 mol of hydrogen. Only 0.3 mol is available, so hydrogen is the limiting reactant and the oxide is in excess.
    3. The hydrogen determines the product: 0.3 mol of H2 gives 0.3 mol of Cu. Mass = 0.3 x 64 = 19.2 g, which is the THEORETICAL yield.
    4. Percentage yield = (actual / theoretical) x 100, so actual = 0.75 x 19.2 = 14.4 g.
    5. Therefore the answer is A.
    • Why not B: Identifies hydrogen as the limiting reactant correctly but then reports the theoretical mass, forgetting to apply the 75% yield at all.
    • Why not C: Takes the copper(II) oxide as limiting because there is more of it by mass, giving 0.5 mol of copper, and then applies the yield to that wrong figure.
    • Why not D: Makes both errors at once: treats the copper(II) oxide as limiting and also omits the yield, so it reports the largest number the numbers can produce.

  2. Question 2Answer: C

    1. Take 100 g of the compound, so the percentages become masses in grams, and divide each by its Ar to get moles: C is 40.0/12 = 3.33, H is 6.7/1 = 6.7 and O is 53.3/16 = 3.33.
    2. Divide through by the smallest, 3.33: C is 1, H is 2 and O is 1. The empirical formula is therefore CH2O.
    3. The empirical formula has a mass of 12 + (2 x 1) + 16 = 30.
    4. The molecular formula is a whole-number multiple of the empirical formula, and that multiple is 180/30 = 6.
    5. Multiplying CH2O by 6 gives C6H12O6, so the answer is C.
    • Why not A: Stops at the empirical formula. It is the correct simplest ratio, but the question asks for the molecular formula and the relative molecular mass of 180 has not been used.
    • Why not B: Scales the empirical formula by 2 rather than 6, which would give a relative molecular mass of 60 rather than the 180 stated in the question.
    • Why not D: Scales the empirical formula by 3, giving a relative molecular mass of 90. This is the value obtained by halving 180, a common slip when the divisor and the dividend are the wrong way round.

  3. Question 3Answer: B

    1. Work in moles first. Moles of sulfuric acid = concentration x volume in dm3 = 0.10 x (20.0/1000) = 0.0020 mol.
    2. The equation shows 1 mol of H2SO4 reacts with 2 mol of NaOH, so moles of NaOH = 2 x 0.0020 = 0.0040 mol.
    3. That amount was contained in 25.0 cm3, so the concentration is 0.0040 / (25.0/1000) = 0.16 mol dm-3.
    4. Convert to g dm-3 using Mr(NaOH) = 23 + 16 + 1 = 40: concentration = 0.16 x 40 = 6.4 g dm-3.
    5. Therefore the answer is B.
    • Why not A: Uses a 1:1 reacting ratio instead of the 1:2 in the equation, so it halves the moles of sodium hydroxide before converting to a mass concentration.
    • Why not C: Applies the 1:2 ratio in the wrong direction, doubling the moles of sodium hydroxide a second time instead of once.
    • Why not D: Reaches the correct concentration in mol dm-3 and then reports that figure with the units of the question, without multiplying by the relative formula mass.

  4. Question 4Answer: A

    1. Assign oxidation states on both sides of each equation and look for one element that both rises and falls.
    2. In A, copper starts as Cu+ at +1. On the right it appears as Cu2+ at +2, which is a rise, and as Cu metal at 0, which is a fall. One element, copper, is both oxidised and reduced.
    3. In B, zinc rises from 0 to +2 and copper falls from +2 to 0, so two different elements change: this is ordinary redox.
    4. In C, hydrogen rises from 0 to +1 and oxygen falls from 0 to -2, again two different elements.
    5. In D, hydrogen stays +1, oxygen stays -2, sodium stays +1 and chlorine stays -1, so nothing is oxidised or reduced.
    6. Only A has one element doing both, so the answer is A.
    • Why not B: This is a redox reaction, but two different elements change: zinc is oxidised from 0 to +2 and copper is reduced from +2 to 0. Disproportionation needs one element doing both.
    • Why not C: Also redox, and again two elements change: hydrogen goes from 0 to +1 and oxygen from 0 to -2. Neither element is both oxidised and reduced.
    • Why not D: No oxidation state changes at all here, so this is not even a redox reaction. It is a neutralisation, which candidates sometimes assume must involve electron transfer because a new compound forms.

  5. Question 5Answer: D

    1. Because the volumes are all measured under the same conditions, they stand in for moles directly, so the reacting ratio is 40 : 50 : 40 for NH3 : O2 : NO.
    2. Divide through by 10 to get 4 : 5 : 4, so the equation must have 4 ammonia, 5 oxygen and 4 nitrogen monoxide.
    3. Complete the balance with water. The left side has 4 x 3 = 12 hydrogen atoms, so 6 H2O are needed on the right.
    4. Check the oxygen: 5 O2 gives 10 oxygen atoms; the right side has 4 in NO and 6 in H2O, which is 10. The equation balances.
    5. That is 4NH3 + 5O2 -> 4NO + 6H2O, so the answer is D.
    • Why not A: Gives an ammonia to oxygen ratio of 2:2, that is 1:1, so it predicts that 40 cm3 of ammonia would need 40 cm3 of oxygen rather than the 50 cm3 measured.
    • Why not B: Has the right ammonia to oxygen ratio of 4:3 for a different reaction, but its product is nitrogen rather than nitrogen monoxide, so it cannot account for the 40 cm3 of NO that was collected.
    • Why not C: Is not balanced: the left side has 3 hydrogen atoms and 2 oxygen atoms while the right side has 2 hydrogen and 2 oxygen, so it cannot represent any real reaction whatever the volumes.

  6. Question 6Answer: B

    1. First find the amount in moles. Mr(NH3) = 14 + (3 x 1) = 17, so moles = 3.4/17 = 0.2 mol.
    2. 0.2 mol of ammonia contains 0.2L molecules, where L is Avogadro's number.
    3. Each NH3 molecule contains one nitrogen atom and three hydrogen atoms, which is four atoms in total.
    4. So the total number of atoms is 0.2L x 4 = 0.8L.
    5. Therefore the answer is B.
    • Why not A: Counts molecules rather than atoms. There are 0.2 mol of ammonia molecules, but the question asks for the total number of atoms, and each molecule contains four of them.
    • Why not C: Counts only the hydrogen atoms, three per molecule, and forgets that the nitrogen atom in each molecule also counts towards the total.
    • Why not D: Treats the mass in grams as though it were a number of moles, so no conversion using the relative molecular mass has been made at all.

  7. Question 7Answer: D

    1. Convert the dissolved mass into moles first. Mr(NaCl) = 23 + 35.5 = 58.5, so moles = 35.1/58.5 = 0.6 mol.
    2. Concentration in mol dm-3 is moles divided by the volume of solution in dm3, and 100 cm3 is 100/1000 = 0.1 dm3.
    3. Concentration = 0.6/0.1 = 6.0 mol dm-3.
    4. Therefore the answer is D.
    • Why not A: Correctly finds the 0.6 mol of sodium chloride present but then quotes that amount as the concentration, without dividing by the 0.1 dm3 of solution it was dissolved in.
    • Why not B: Scales in the wrong direction. Going from a 100 cm3 sample to 1 dm3 means multiplying the mass by 10, not dividing it, so this figure is 35.1/10 rather than the 351 g dm-3 that scaling actually gives. The mass has not been converted into moles either, so the number is a mass concentration wearing the units of a molar one.
    • Why not C: Treats 100 cm3 as 10 dm3 instead of 0.1 dm3, so it divides the 0.6 mol by 10 rather than by 0.1. Moving the decimal point the wrong way when converting cm3 to dm3 is the commonest slip in this calculation, and it makes the answer a hundred times too small.

  8. Question 8Answer: C

    1. Write the mass fraction of M in M2O3 in terms of the unknown Ar. The formula contains two M atoms and three oxygen atoms, so the relative formula mass is 2M + (3 x 16) = 2M + 48.
    2. The metal accounts for 70% of that mass, so 2M/(2M + 48) = 0.70.
    3. Multiply out: 2M = 0.70(2M + 48) = 1.4M + 33.6.
    4. Collect the terms in M: 2M - 1.4M = 33.6, so 0.6M = 33.6 and M = 56.
    5. Check it: 2 x 56 = 112 and 112 + 48 = 160, and 112/160 = 0.70 as required. Therefore the answer is C.
    • Why not A: Is the relative atomic mass of aluminium, whose oxide is also M2O3. It is a reasonable guess at the identity of the metal but it does not satisfy the 70% figure, which gives 52.9% aluminium.
    • Why not B: Sets up the correct equation 2M = 0.70(2M + 48) and correctly reaches 0.6M = 33.6, but then divides 33.6 by 0.70 instead of 0.6, giving 48 rather than the correct 56.
    • Why not D: Is twice the correct value, obtained by solving for the mass of the two metal atoms together, 2M, and then reporting that total as the relative atomic mass of a single atom.

  9. Question 9Answer: A

    1. An increase in overall pressure shifts an equilibrium towards whichever side has FEWER moles of gas, because that reduces the pressure again.
    2. Count the moles of gas on each side of 2HI(g) <=> H2(g) + I2(g). The left has 2 mol of gas. The right has 1 mol of H2 plus 1 mol of I2, which is also 2 mol of gas.
    3. The two sides are equal, so neither side is favoured and there is nothing for the system to relieve by shifting.
    4. The position of equilibrium therefore does not move. Note that the endothermic label is there to be set aside: it governs the response to temperature, not to pressure.
    5. Therefore the answer is A.
    • Why not B: Counts substances rather than moles of gas. Two products are formed but they total 2 mol of gas, exactly matching the 2 mol on the left, so there is no side for the equilibrium to favour.
    • Why not C: States the pressure rule correctly but does not check it against this equation. Neither side has fewer molecules of gas here, so the rule has nothing to act on.
    • Why not D: Confuses two separate factors. Pressure and temperature act independently, and in any case a lower temperature would favour the exothermic direction rather than being a restatement of a pressure change.

  10. Question 10Answer: D

    1. Assign oxidation states on both sides. In Fe2O3 oxygen is -2, so the two iron atoms must total +6 and each iron is +3. In elemental carbon the state is 0, because it is an element in its standard form.
    2. On the right, iron is a pure metal, so it is 0. In CO2 oxygen is -2, giving -4 for the two oxygens, so carbon must be +4.
    3. Iron therefore falls from +3 to 0, which is reduction. Carbon rises from 0 to +4, which is oxidation.
    4. The reducing agent is the species that causes the reduction by being oxidised itself, and that is the carbon.
    5. Therefore the answer is D.
    • Why not A: Picks the right formula for the wrong role and reverses the direction. Iron in Fe2O3 starts at +3, not 0, and it falls rather than rises.
    • Why not B: Describes what happens to the iron correctly, but a species that is reduced is the OXIDISING agent. The reducing agent is the one that is itself oxidised.
    • Why not C: Names a product rather than a reactant. An agent must be something present at the start, and carbon in CO2 has already finished at +4 rather than changing to it.

  11. Question 11Answer: B

    1. Find the relative formula mass of both solids. For CuSO4: 64 + 32 + (4 x 16) = 160. Each water molecule is 18, so the hydrate CuSO4.5H2O is 160 + (5 x 18) = 160 + 90 = 250.
    2. Convert the starting mass to moles: 25/250 = 0.1 mol of the hydrate.
    3. Every mole of hydrate leaves one mole of anhydrous CuSO4 behind, so 0.1 mol of CuSO4 remains.
    4. Mass = 0.1 x 160 = 16.0 g.
    5. As a check, the water lost is 0.1 x 90 = 9.0 g, and 16.0 + 9.0 = 25.0 g as it must. Therefore the answer is B.
    • Why not A: Calculates the mass of water driven off rather than the solid that remains. It is the right subtraction done from the wrong end of the question.
    • Why not C: Notices the formula has 5 water molecules and assumes water must therefore make up one-fifth of the total mass, so treats 25/5 = 5.0 g as the mass lost and reports 25 - 5.0 = 20.0 g as what remains, without ever finding a relative formula mass.
    • Why not D: Assumes the mass does not change on heating. Mass is conserved overall, but the water leaves as a vapour, so the solid that remains must be lighter.

  12. Question 12Answer: C

    1. Only the calcium carbonate decomposes, so find its mass first: 80% of 12.5 g = 0.80 x 12.5 = 10.0 g.
    2. Convert that to moles. Mr(CaCO3) = 40 + 12 + (3 x 16) = 100, so moles = 10.0/100 = 0.10 mol.
    3. The equation is 1:1, so 0.10 mol of calcium carbonate gives 0.10 mol of carbon dioxide.
    4. Volume = moles x molar volume = 0.10 x 24 = 2.4 dm3.
    5. Therefore the answer is C.
    • Why not A: Applies the 80% purity figure twice: correctly finds the mass of calcium carbonate as 0.80 x 12.5 = 10.0 g and the gas volume that gives, 10.0/100 x 24 = 2.4 dm3, but then takes 80% of that volume too, as though the percentage needed to be used a second time, giving 2.4 x 0.80 = 1.92 dm3.
    • Why not B: Ignores the purity entirely and uses all 12.5 g as calcium carbonate, so it overstates the gas by the fraction of the sample that cannot decompose.
    • Why not D: Stops at the number of moles of carbon dioxide and reports it as a volume, without multiplying by the molar volume of 24 dm3.

  13. Question 13Answer: D

    1. The mass of nitrogen that combined is the difference between the product and the magnesium: 1.00 - 0.72 = 0.28 g.
    2. Convert each mass to moles. Magnesium: 0.72/24 = 0.03 mol. Nitrogen: 0.28/14 = 0.02 mol.
    3. The ratio of moles is Mg : N = 0.03 : 0.02.
    4. Divide both by the smaller, 0.02, to get 1.5 : 1, then multiply both by 2 to clear the fraction, giving 3 : 2.
    5. The formula is therefore Mg3N2, so the answer is D.
    • Why not A: Comes from assuming a 1:1 ratio because the compound has two elements, without dividing either mass by its relative atomic mass to compare amounts.
    • Why not B: Has the two subscripts the right way round for the numbers 2 and 3 but attached to the wrong elements, which is what happens if the mole ratio is written as N:Mg and then read as Mg:N.
    • Why not C: Carries the diatomic formula of the element straight into the compound: nitrogen gas is N2, so the compound is assumed to contain an N2 unit. A subscript in a formula counts atoms of that element in the compound, not atoms in the molecule the element arrived as.

  14. Question 14Answer: A

    1. Find the moles of acid used: 0.10 x (20.0/1000) = 0.0020 mol of HCl.
    2. The equation shows 2 mol of HCl react with 1 mol of M(OH)2, so moles of hydroxide = 0.0020/2 = 0.0010 mol.
    3. That was contained in 25.0 cm3, so the concentration is 0.0010/(25.0/1000) = 0.040 mol dm-3.
    4. The same solution is 2.32 g dm-3, and concentration in g dm-3 = concentration in mol dm-3 x Mr, so Mr = 2.32/0.040 = 58.
    5. The formula M(OH)2 contains two OH groups, each of mass 16 + 1 = 17, so Ar(M) = 58 - (2 x 17) = 58 - 34 = 24.
    6. Therefore the answer is A.
    • Why not B: Reaches the correct relative formula mass of 58 for the hydroxide but confuses one hydroxide group with a whole water molecule and subtracts 18 instead of the 34 that the two OH groups in M(OH)2 actually contribute: 58 - 18 = 40.
    • Why not C: Quotes the relative formula mass of the whole hydroxide rather than the relative atomic mass of the metal alone, so the two hydroxide groups have never been removed.
    • Why not D: Is the relative atomic mass of copper. Cu(OH)2 is a real, commonly met hydroxide, but it does not fit the data given: at the titration's own concentration of 0.040 mol dm-3 it would have a mass concentration of 0.040 x 98 = 3.92 g dm-3, not the 2.32 g dm-3 stated.

  15. Question 15Answer: C

    1. Count the moles of gas on each side. The left has 1 mol of CO plus 2 mol of H2, which is 3 mol of gas. The right has 1 mol of CH3OH.
    2. Raising the overall pressure shifts an equilibrium towards the side with fewer moles of gas, which here is the methanol side, so the yield rises. That makes C correct.
    3. Checking the others: a catalyst changes the rate of both directions equally and so cannot move the position of an equilibrium at all.
    4. Raising the temperature favours the endothermic direction, and since the forward reaction is exothermic that is the reverse reaction, lowering the yield.
    5. Removing carbon monoxide takes away a reactant, so the equilibrium shifts left to replace it, again lowering the yield.
    6. Therefore the answer is C.
    • Why not A: A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached sooner but its position, and therefore the yield, is unchanged.
    • Why not B: Reverses the temperature rule. The forward reaction is exothermic, so raising the temperature shifts the equilibrium towards the endothermic reverse reaction and lowers the yield.
    • Why not D: Removes a reactant, which shifts the equilibrium to the left to replace it. Removing the PRODUCT would raise the yield, but taking away carbon monoxide lowers it.

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