ESAT Chemistry Practice Questions by Topic

Correct Answer: B. $16$

Fastest Approach (🚀):
Hydrogen has only one place to go, so halve its 24 atoms and read the water count straight off as 12; nitrogen is 8 halved, and the answer is the sum.

Step-by-Step Breakdown:

1. Fix what the spark cannot change.

A chemical change rearranges atoms and their electrons: bonds break and re-form, but no nucleus is created or destroyed. The flask is sealed, so it holds the same 24 hydrogen, 8 nitrogen and 12 oxygen atoms after the spark as before it.

2. Place the hydrogen.

Among the products, hydrogen appears only in water, and each water molecule takes two hydrogen atoms. The 24 hydrogen atoms therefore give $24÷2=12$ molecules of water.

3. Check the oxygen.

Those 12 water molecules take one oxygen atom each, which is 12 oxygen atoms, and exactly 12 are present. Every oxygen atom is used up, so no oxygen molecule is left in the flask.

4. Place the nitrogen and total up.

Nitrogen appears in the products only as ${\mathrm{N}}_2$, two atoms per molecule, so the 8 nitrogen atoms give $8÷2=4$ molecules. The flask then holds $12+4=16$ molecules.

The 16 product molecules carry $12\times 3=36$ atoms in the water and $4\times 2=8$ atoms in the nitrogen, which is 44 atoms in all, the same total as the $24+8+12=44$ atoms the flask started with.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $14$ · Conceptual Error
  The molecule count before the spark: 8 nitrogen atoms sit in 8 ammonia molecules and 12 oxygen atoms in 6 oxygen molecules, so $8+6=14$. Atoms are conserved by a reaction, but molecules are made and unmade, so 14 need not survive.
• C. $20$ · Miscount
  The water is right at 12, but the nitrogen is then counted as 8 loose atoms instead of 4 molecules of ${\mathrm{N}}_2$, giving $12+8=20$.
• D. $22$ · Unchecked Assumption
  12 water and 4 nitrogen molecules, plus 6 oxygen molecules assumed to be left over: $12+4+6=22$. The 12 water molecules take all 12 oxygen atoms, so no oxygen remains.
• E. $28$ · Formula Error
  One water molecule per hydrogen atom gives 24, and the 4 nitrogen molecules bring it to $24+4=28$. Water is ${\mathrm{H}}_2\mathrm{O}$, so each molecule takes two hydrogen atoms, not one.
• F. $32$ · Wrong Quantity Counted
  The atoms supplied by the ammonia: 8 molecules of 4 atoms each, $8\times 4=32$. That counts atoms going into the reaction rather than molecules coming out of it.
• G. $36$ · Atoms Counted As Molecules
  The atoms inside the water, $12\times 3=36$, reported in place of the 12 water molecules, with the nitrogen left out altogether.
• H. $44$ · Atoms Counted As Molecules
  Every atom in the flask, $24+8+12=44$. That total genuinely is unchanged by the spark, but the question asks how many molecules there are.

Common Mistake (⚠️):
Carrying conservation too far and applying it to molecules. The flask holds 14 molecules before the spark, 8 of ammonia and 6 of oxygen, and a candidate who assumes that number is protected answers 14. Nuclei and atoms survive a reaction; molecules are taken apart and rebuilt.

Takeaway (📌):
In a sealed vessel, treat the atoms of each element as fixed, never the molecules. Share each element out among the product formulae and the molecule totals fall out of the division.

Correct Answer: B. (s), (aq), (aq), (l), (g)

Fastest Approach (🚀):
Settle the acid and the water first, since those two carry the most information here. Hydrochloric acid on the bench is a solution, so it is (aq), and the water made is the pure liquid, so it is (l). Fixing those two positions narrows the row before you look at anything else.

Step-by-Step Breakdown:

1. Take the five species in the order listed and attach the symbol the description demands.

$\mathrm{C}\mathrm{a}\mathrm{C}{\mathrm{O}}_3$: the chips are stated to stay undissolved, so the carbonate is present as solid throughout the reaction and takes (s).
$\mathrm{H}\mathrm{C}\mathrm{l}$: bench hydrochloric acid is hydrogen chloride dissolved in water, not a pure liquid, so it takes (aq). The symbol (l) is reserved for a substance that is both liquid and pure, which this is not.
$\mathrm{C}\mathrm{a}\mathrm{C}{\mathrm{l}}_2$: the calcium chloride is stated to stay in solution, so it is a solute in water and takes (aq).
${\mathrm{H}}_2\mathrm{O}$: the water formed is the pure liquid itself. It is the solvent that (aq) refers to, so it takes (l).
$\mathrm{C}{\mathrm{O}}_2$: the fizzing is carbon dioxide leaving the mixture as bubbles, so it takes (g).
Written out, the equation is $$\mathrm{C}\mathrm{a}\mathrm{C}{\mathrm{O}}_3(\mathrm{s})+2\mathrm{H}\mathrm{C}\mathrm{l}(\mathrm{a}\mathrm{q})$$$$\to \mathrm{C}\mathrm{a}\mathrm{C}{\mathrm{l}}_2(\mathrm{a}\mathrm{q})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$$$+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$$, so the row of symbols is (s), (aq), (aq), (l), (g).

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. (s), (l), (aq), (l), (g) · Conceptual Error
  Marks the acid (l) because it is poured as a liquid. Hydrochloric acid is hydrogen chloride dissolved in water, so it is (aq); (l) would mean pure liquid hydrogen chloride, which is not what is in the bottle.
• C. (s), (l), (l), (aq), (g) · Conceptual Error
  Has the two symbols the wrong way round, writing (l) for the substances dissolved in the water and (aq) for the water itself. (aq) marks a solute in water and (l) marks a pure liquid, so the acid and the salt are (aq) and the water is (l).
• D. (s), (l), (l), (l), (g) · Conceptual Error
  Never uses (aq) at all, marking (l) for everything present in the liquid mixture, so both the acid and the salt come out as (l). A substance dissolved in water takes (aq), and (l) belongs only to the pure liquid, which here is the water.
• E. (s), (l), (s), (l), (g) · Conceptual Error
  Labels each substance by what a pure sample of it looks like on the shelf: a solid carbonate, an acid poured from a bottle, a jar of solid calcium chloride. State symbols describe what is in the flask, where the acid and the salt are both dissolved in water, so both take (aq).
• F. (aq), (aq), (aq), (l), (aq) · Misreading the Stem
  Marks the chips and the carbon dioxide (aq) on the idea that anything sharing the flask with the solution counts as dissolved. The stem says the chips stay undissolved, so they are (s), and the fizzing is gas escaping the mixture, so the carbon dioxide is (g).

Common Mistake (⚠️):
Writing $\mathrm{H}\mathrm{C}\mathrm{l}(\mathrm{l})$ for the acid because it is poured from a bottle as a liquid. Bench hydrochloric acid is hydrogen chloride dissolved in water, so it takes (aq), while (l) belongs to a substance that is liquid and pure, such as the water this reaction makes.

Takeaway (📌):
Ask of every species, is it dissolved in water? If it is, the symbol is (aq), whatever the bottle looked like. Water itself is the solvent, so water produced in a reaction in aqueous solution is written (l) and never (aq).

Correct Answer: D. $1.06\text{ g}$

Fastest Approach (🚀):
Keep the powers of ten together and the $M_r$ until last. $0.100\times 20.0{\text{ cm}}^3$ gives $2\times {10}^{-3}\text{ mol}$ of acid, halving for the ratio gives ${10}^{-3}$, and the factor of ten for the full flask restores ${10}^{-2}$, so the mass is simply $106÷100$.

Step-by-Step Breakdown:

1. Find the moles of acid delivered from the burette

$$n(\mathrm{H}\mathrm{C}\mathrm{l})=c\times V$$$$=0.100\times {\displaystyle \frac{20.0}{1000}}$$$$=2.00\times {10}^{-3}\text{ mol}$$

2. Convert to moles of carbonate in the portion titrated

The equation uses $2$ mol of $\mathrm{H}\mathrm{C}\mathrm{l}$ for every $1$ mol of $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$, so

$$n(\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3)={\displaystyle \frac{2.00\times {10}^{-3}}{2}}$$$$=1.00\times {10}^{-3}\text{ mol}$$

This is the amount in the $25.0{\text{ cm}}^3$ portion only.

3. Scale the portion up to the whole flask

$25.0{\text{ cm}}^3$ is one tenth of $250{\text{ cm}}^3$, so the flask held

$n=10\times 1.00\times {10}^{-3}=1.00\times {10}^{-2}\text{ mol}$

4. Convert moles to mass

$m=n\times M_r=1.00\times {10}^{-2}\times 106=1.06\text{ g}$

Sanity check: about one hundredth of a mole of a substance with $M_r$ close to $100$ should weigh close to $1\text{ g}$, which it does.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $0.106\text{ g}$ · Aliquot not scaled up
  The mole ratio was applied correctly to give $1.00\times {10}^{-3}\text{ mol}$ in the portion, but that was converted straight to mass: $1.00\times {10}^{-3}\times 106=0.106\text{ g}$. This is the mass in the $25.0{\text{ cm}}^3$ portion, one tenth of what was weighed out.
• B. $0.212\text{ g}$ · Ratio and dilution both ignored
  Both conversions were missed: the acid moles were used as the carbonate moles and the portion was never scaled up, giving $2.00\times {10}^{-3}\times 106=0.212\text{ g}$.
• C. $0.530\text{ g}$ · Mole ratio applied twice
  The factor of two was applied twice, once to the acid moles and then again before scaling up: $2.00\times {10}^{-3}÷2÷2=5.00\times {10}^{-4}$, then $\times 10\times 106=0.530\text{ g}$. The ratio is used once only.
• E. $2.12\text{ g}$ · Mole ratio ignored
  The scale up to $250{\text{ cm}}^3$ was done correctly but the $2:1$ ratio was ignored: $2.00\times {10}^{-3}\times 10\times 106=2.12\text{ g}$, exactly double the true mass.

Common Mistake (⚠️):
Treating the moles of hydrochloric acid run in from the burette as the moles of sodium carbonate present. The equation needs two acid molecules per carbonate, so the carbonate amount is half the acid amount, and carrying $2.00\times {10}^{-3}\text{ mol}$ forward instead of $1.00\times {10}^{-3}\text{ mol}$ doubles every line after it.

Takeaway (📌):
Every titration question hides two conversion factors, not one. Before touching $M_r$, ask what the mole ratio is and what fraction of the solution was actually in the flask, then apply both.

Correct Answer: C. $160{\text{ cm}}^3$

Fastest Approach (🚀):
$$1\text{ vol }{\mathrm{C}}_3{\mathrm{H}}_8+5\text{ vol }{\mathrm{O}}_2$$$$\to 3\text{ vol }\mathrm{C}{\mathrm{O}}_2+4\text{ vol }{\mathrm{H}}_2\mathrm{O}$$.
$20{\text{ cm}}^3$ gives $60{\text{ cm}}^3$ of $\mathrm{C}{\mathrm{O}}_2$ and $80{\text{ cm}}^3$ of ${\mathrm{H}}_2\mathrm{O}$.
Oxygen used $=5\times 20=100{\text{ cm}}^3$. Remaining ${\mathrm{O}}_2=120-100=20{\text{ cm}}^3$.
Total gas $=60+80+20=160{\text{ cm}}^3$. Matches Option C.

Step-by-Step Breakdown:

1. Write the balanced equation

$${\mathrm{C}}_3{\mathrm{H}}_8(\mathrm{g})+5{\mathrm{O}}_2(\mathrm{g})$$$$\to 3\mathrm{C}{\mathrm{O}}_2(\mathrm{g})+4{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$$

2. Determine the limiting reactant

By Avogadro's law, gas volumes react in the same ratio as their moles.
$20{\text{ cm}}^3$ of ${\mathrm{C}}_3{\mathrm{H}}_8$ requires $5\times 20=100{\text{ cm}}^3$ of ${\mathrm{O}}_2$.
Since we have $120{\text{ cm}}^3$ of ${\mathrm{O}}_2$, ${\mathrm{C}}_3{\mathrm{H}}_8$ is the limiting reactant and ${\mathrm{O}}_2$ is in excess.

3. Calculate volumes of products formed

Volume of $\mathrm{C}{\mathrm{O}}_2=3\times 20=60{\text{ cm}}^3$.
Volume of ${\mathrm{H}}_2\mathrm{O}(\mathrm{g})=4\times 20=80{\text{ cm}}^3$.

4. Calculate excess reactant remaining

Excess ${\mathrm{O}}_2=120-100=20{\text{ cm}}^3$.

5. Sum all final gases

Total volume $$=V(\mathrm{C}{\mathrm{O}}_2)+V({\mathrm{H}}_2\mathrm{O})$$$$+V({\mathrm{O}}_2\text{ remaining})$$
Total volume $=60+80+20=160{\text{ cm}}^3$.
Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $60{\text{ cm}}^3$ · Only one product counted
  This is the carbon dioxide on its own, $3\times 20=60{\text{ cm}}^3$. The question says the water stays a gas, so it counts too, and the unused oxygen is still in the vessel.
• B. $140{\text{ cm}}^3$ · Omission Error
  Calculated the volume of products ($60+80=140$) but forgot to add the unreacted excess oxygen ($20$).
• D. $120{\text{ cm}}^3$ · A figure from the question, not an answer
  $120{\text{ cm}}^3$ is the oxygen supplied, which is given in the question. It is also the volume that reacts, $20$ of propane with $100$ of oxygen, so it measures what disappears rather than what is there at the end.
• E. $100{\text{ cm}}^3$ · Intermediate value
  $5\times 20=100{\text{ cm}}^3$ is the oxygen the propane consumes. That is a step on the way to the answer, used to show oxygen is in excess, not the gas remaining at the end.

Common Mistake (⚠️):
Forgetting to include the unreacted oxygen in the final total volume, leading to $140{\text{ cm}}^3$.

Takeaway (📌):
Balance, find the limiting reactant, then count everything left in the vessel and not only what was made. The leftover excess is the part most often dropped, and here it is the whole difference between $140{\text{ cm}}^3$ and the right answer.

Correct Answer: B. 0.672

Fastest Approach (🚀):
$96\mathrm{c}{\mathrm{m}}^3$ out of the $24000\mathrm{c}{\mathrm{m}}^3$ in one mole is $0.004\mathrm{m}\mathrm{o}\mathrm{l}$ of carbon dioxide; scale by $\frac{2}{1}$ to get $0.008\mathrm{m}\mathrm{o}\mathrm{l}$, then multiply by $84$. Two options go without any arithmetic: whichever is a thousand times another is the $\mathrm{c}{\mathrm{m}}^3$ against $\mathrm{d}{\mathrm{m}}^3$ slip, and $0.008$ on its own is an amount in $\mathrm{m}\mathrm{o}\mathrm{l}$, not a mass.

Step-by-Step Breakdown:

1. Recover the total volume of carbon dioxide

A mean rate is the total change divided by the time it took, so the total is the rate multiplied by the interval. Only the carbon dioxide reaches the syringe, so its reading follows that one substance:

$$V=1.2\times 80=96\mathrm{c}{\mathrm{m}}^3$$$$=0.096\mathrm{d}{\mathrm{m}}^3$$

2. Convert to moles of carbon dioxide, then to moles of sodium hydrogencarbonate

$$n(\mathrm{C}{\mathrm{O}}_2)=\frac{0.096}{24}$$$$=0.004\mathrm{m}\mathrm{o}\mathrm{l}$$

The balanced equation pairs $2\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}{\mathrm{O}}_3$ with $1\mathrm{C}{\mathrm{O}}_2$, so the amount of sodium hydrogencarbonate is $\frac{2}{1}$ times the amount of carbon dioxide:

$$n(\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}{\mathrm{O}}_3)=\frac{2}{1}\times 0.004$$$$=0.008\mathrm{m}\mathrm{o}\mathrm{l}$$

3. Convert the amount to a mass

$$m=n\times M_r=0.008\times 84$$$$=0.672g$$

Check the units before committing. Dividing $96\mathrm{c}{\mathrm{m}}^3$ by a molar volume quoted in $\mathrm{d}{\mathrm{m}}^3$ would have given $4\mathrm{m}\mathrm{o}\mathrm{l}$ of carbon dioxide and a final mass of $672g$, which a syringe holding $96\mathrm{c}{\mathrm{m}}^3$ cannot account for.

The key is $0.672$.

Why the Other Options Are Wrong (❌):

• A. 0.008 · Amount in moles reported as a mass
  The amount in moles was written down as though it were already a mass: $n(\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}{\mathrm{O}}_3)=0.008\mathrm{m}\mathrm{o}\mathrm{l}$ was quoted as $0.008g$. The last conversion is still owed, and it is the one the stem hands over: multiplying by $M_r(\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}{\mathrm{O}}_3)=84$ gives $0.672$.
• C. 0.336 · Stoichiometric ratio ignored
  The equation's coefficients were skipped and the amount of sodium hydrogencarbonate was set equal to the amount of carbon dioxide, $0.004\mathrm{m}\mathrm{o}\mathrm{l}$ rather than $0.008\mathrm{m}\mathrm{o}\mathrm{l}$, before multiplying by $84$. The equation pairs $2\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}{\mathrm{O}}_3$ with $1\mathrm{C}{\mathrm{O}}_2$, so the amount of sodium hydrogencarbonate is $\frac{2}{1}$ times the amount of carbon dioxide, and this option is the key $0.672$ scaled by $\frac{1}{2}$.
• D. 0.848 · Wrong relative formula mass used
  The right amount, $0.008\mathrm{m}\mathrm{o}\mathrm{l}$, was multiplied by the relative formula mass of sodium carbonate instead: $106$ rather than the $84$ the stem supplies. $M_r(\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3)=106$ is genuine, but it belongs to another substance in the equation, and the question asks for the mass of sodium hydrogencarbonate.
• E. 672 · Molar volume units mismatched
  The volume was left in $\mathrm{c}{\mathrm{m}}^3$ and divided by a molar volume quoted in $\mathrm{d}{\mathrm{m}}^3$: $96÷24=4\mathrm{m}\mathrm{o}\mathrm{l}$ of carbon dioxide instead of $0.004\mathrm{m}\mathrm{o}\mathrm{l}$. One mole of gas occupies $24\mathrm{d}{\mathrm{m}}^3$, which is $24000\mathrm{c}{\mathrm{m}}^3$, so this option is the key $0.672$ multiplied by a thousand.

Common Mistake (⚠️):
Starting the mole calculation from the rate itself. A figure in $\mathrm{c}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ is not a volume, so it has to be multiplied by the $80s$ interval before the molar gas volume can be used; feeding $1.2$ straight into the division gives an amount $80$ times too small.

Takeaway (📌):
Rate times time gives the volume of carbon dioxide; the molar gas volume, the $2$ to $1$ from the balanced equation and $M_r(\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}{\mathrm{O}}_3)=84$ then run the chain backwards to the sodium hydrogencarbonate. Following a product is only ever a way of following the reactant that made it.

Correct Answer: E. $208{\text{ kJ mol}}^{-1}$

Fastest Approach (🚀):
Subtract $\Delta H$ from the forward barrier: $268-(+60)=208$. Matches Option E.

Step-by-Step Breakdown:

1. Picture the reaction profile

The forward activation energy is measured from the reactants up to the transition state; the reverse activation energy is measured from the products up to that same transition state.

2. Relate the two barriers

$$\Delta H=E_{a,\text{fwd}}-E_{a,\text{rev}}$$$$⟹E_{a,\text{rev}}=E_{a,\text{fwd}}-\Delta H$$

3. Substitute

$$E_{a,\text{rev}}=268-(+60)$$$$=208{\text{kJ mol}}^{-1}$$

4. Check it is physical

The reaction is endothermic ($\Delta H=+60$ kJ mol${}^{-1}$), so the reverse barrier must be smaller than the forward one, and $208<268$. Both barriers are positive, as every activation energy must be.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $60{\text{ kJ mol}}^{-1}$ · Confused quantities
  $60{\text{ kJ mol}}^{-1}$ is the enthalpy change, the height difference between reactants and products. An activation energy is the height of the barrier above one of them, which is a different measurement on the same diagram.
• B. $416{\text{ kJ mol}}^{-1}$ · Impossible on the reaction profile
  $416$ is larger than the forward barrier of $268$, which cannot happen when the products start higher than the reactants, and no single sum or difference of $268$ and $60$ produces it either.
• C. $268{\text{ kJ mol}}^{-1}$ · Answered the wrong quantity
  $268{\text{ kJ mol}}^{-1}$ is the forward activation energy, which the question supplies. The reverse barrier is the one being asked for.
• D. $328{\text{ kJ mol}}^{-1}$ · Sign error
  $268+60=328$: $\Delta H$ added rather than subtracted. That would put the reverse barrier above the forward one, and for an endothermic reaction it has to be below.

Common Mistake (⚠️):
Adding the enthalpy change directly to the forward activation energy regardless of sign.

Takeaway (📌):
In an endothermic reaction the products sit above the reactants, so the climb from the product side up to the transition state is the shorter one: the reverse barrier must come out smaller than the forward barrier. That alone removes every option at or above $268{\text{ kJ mol}}^{-1}$ before any arithmetic.

Correct Answer: A. 12

Fastest Approach (🚀):
Compute $mc=200\times 4.2=840\text{J}$ per degree first and write the heat straight in joules as $10080$. Faster still, the mass in grams is numerically equal to the volume in ${\text{cm}}^3$, so the volume cancels and the rise is just $0.2\times 252÷4.2=12$, with no litres and no grams to carry.

Step-by-Step Breakdown:

1. Find the amount of manganese displaced

The magnesium is in excess, so every manganese ion reacts and the manganese(II) sulfate is the limiting reagent. The volume in cubic decimetres is $0.2{\text{dm}}^3$, so

$$n(\mathrm{M}\mathrm{n}\mathrm{S}{\mathrm{O}}_4)=c\times V=0.2\times 0.2$$$$=0.04\text{mol}$$

The reaction is $\mathrm{M}\mathrm{g}+\mathrm{M}{\mathrm{n}}^{2+}\to \mathrm{M}{\mathrm{g}}^{2+}+\mathrm{M}\mathrm{n}$, one manganese atom for each manganese ion, so $0.04\text{mol}$ of manganese is displaced.

2. Convert that amount into a total heat

$$Q=0.04\times 252=10.08\text{kJ}$$$$=10080\text{J}$$

3. Turn the heat into a temperature rise

What warms up is the $200{\text{cm}}^3$ of solution, and at a density of $1{\text{g cm}}^{-3}$ that is $200\text{g}$. The metal's mass is ignored and the cup loses no heat, so all $10080\text{J}$ goes into the solution. Rearranging $Q=mc\Delta T$,

$$\Delta T=\frac{Q}{mc}=\frac{10080}{200\times 4.2}$$$$=\frac{10080}{840}=12{}^{\circ }\text{C}$$

Sanity check: $840\text{J}$ raises this solution by one degree, and $10080$ is $12$ times $840$, so $12$ degrees. Notice that the volume never reaches the answer: it fixes the amount reacting and the mass warmed in the same proportion, so it cancels, and the rise depends only on the concentration, the energy per mole and the specific heat capacity.

The temperature rise is $12{}^{\circ }\text{C}$.

Why the Other Options Are Wrong (❌):

• B. 1.2 · Volume conversion off by ten
  The volume was converted the wrong way, $200{\text{cm}}^3$ taken as $0.02{\text{dm}}^3$, giving $n=0.004\text{mol}$ and $Q=1.008\text{kJ}$, then $1008÷840=1.2$. A ${\text{dm}}^3$ is a thousand ${\text{cm}}^3$, so the volume is $0.2{\text{dm}}^3$ and the amount reacting is ten times larger than this.
• C. 6 · Ionic charge used as a mole ratio
  The $2+$ charge on the manganese ion was used as a stoichiometric factor and the amount divided by $2$ to $0.02\text{mol}$: $0.02\times 252=5.04\text{kJ}$ and $5040÷840=6$. The charge counts the electrons transferred, not the atoms: each manganese ion gives one manganese atom, so the amount stays at $0.04\text{mol}$.
• D. 50.4 · Specific heat capacity omitted
  The specific heat capacity was dropped from the denominator: $10080÷200=50.4$. Dividing joules by a mass alone leaves ${\text{J g}}^{-1}$, an energy per gram and not a temperature at all; it is the $4.2$ that turns joules per gram into degrees.
• E. 300 · Scaling by the amount omitted
  The per-mole figure was treated as the total heat released: $252\text{kJ}=252000\text{J}$ and $252000÷840=300$. Only $0.04\text{mol}$ of manganese is displaced, so only that fraction of the $252\text{kJ}$ is ever given out, and $300$ is the rise a whole mole would produce.

Common Mistake (⚠️):
Dividing the heat by the amount instead of multiplying, out of habit from the more familiar calorimetry question that asks for kilojoules per mole. Here the energy per mole is handed over, $252\text{kJ}$, and the temperature rise is what is wanted, so the amount $0.04\text{mol}$ multiplies up to a total heat of $10080\text{J}$ first, and only then does $Q=mc\Delta T$ get rearranged.

Takeaway (📌):
Work out what one degree costs before anything else. The product $mc$ is the price of a degree in joules, here $840\text{J}$, and every calorimetry question is then either a heat divided by that price or that price multiplied by a rise.

Correct Answer: C. $200{\text{ kJ mol}}^{-1}$

Fastest Approach (🚀):
$\Delta H=4000-3800=200$. Matches Option C.

Step-by-Step Breakdown:

1. Write the rule down

$$\Delta H$$$$=\text{energy to break the reactant bonds}$$$$-\text{energy released making the product bonds}$$

Breaking bonds costs energy and making them releases it, so the first total is what the reaction takes in and the second is what it gives back.

2. Substitute the two totals

$\Delta H=4000-3800=+200{\text{ kJ mol}}^{-1}$

3. Read the sign

More energy was spent breaking bonds than was recovered making them, so the reaction takes $200{\text{ kJ mol}}^{-1}$ from its surroundings: $\Delta H$ is positive and the reaction is endothermic. A mixture that cools as it reacts is the same statement seen from the outside.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $-7800{\text{ kJ mol}}^{-1}$ · Totals added, then negated
  $4000+3800=7800$, made negative. Adding the two totals measures the energy of every bond in the reaction; the enthalpy change is the difference between the two sides.
• B. $-2127{\text{ kJ mol}}^{-1}$ · Not derivable from the data
  $-2127$ follows from no route through these two numbers. The totals differ by $200$, so $\Delta H$ has magnitude $200$ however the subtraction is arranged, and a glance at the size removes this option before any arithmetic.
• D. $-200{\text{ kJ mol}}^{-1}$ · Sign Error
  Reversed the subtraction (Made - Broken), leading to the wrong sign.
• E. $7800{\text{ kJ mol}}^{-1}$ · Totals added
  $4000+3800=7800$: the two totals added instead of subtracted. That figure is the energy of all the bonds involved, which no reaction ever absorbs or releases.

Common Mistake (⚠️):
Subtracting the broken bonds from the made bonds, getting the wrong sign.

Takeaway (📌):
Broken minus made, in that order, and then read the sign instead of discarding it. The two totals here differ by $200$, so the answer has magnitude $200$ whichever way round the subtraction is done, and every option that is not $\pm 200$ can go before a line of working is written.