A Level H1 Chemistry Stoichiometry Moles Quiz

A-Level Chemistry H1 Quiz - Stoichiometry Moles

Answer Key

---

Section A: Multiple Choice

1. C [2 marks]

Explanation: The formula $Ca(NO_3)_2$ contains 1 Ca, 2 N, and 6 O atoms per formula unit. In 0.50 mol of $Ca(NO_3)_2$, the number of moles of oxygen atoms = $0.50 \times 6 = 3.0$ mol. Students often mistakenly count only 3 oxygen atoms (forgetting the subscript 2 outside the bracket) and choose B (1.0 mol) or A (0.50 mol).

---

2. C [2 marks]

Explanation: The empirical formula mass of $CH_2O = 12.0 + 2(1.0) + 16.0 = 30.0$ g mol$^{-1}$. The multiple = $180 / 30 = 6$. Therefore, the molecular formula = $(CH_2O)_6 = C_6H_{12}O_6$. This is glucose. A common error is to divide incorrectly or to stop at the empirical formula.

---

3. A [2 marks]

Explanation: The number of moles = mass / molar mass. For 1.0 g of each:

• $H_2$: $1.0 / 2.0 = 0.50$ mol
• $O_2$: $1.0 / 32.0 = 0.031$ mol
• $N_2$: $1.0 / 28.0 = 0.036$ mol
• $CO_2$: $1.0 / 44.0 = 0.023$ mol

Since $H_2$ has the smallest molar mass, 1.0 g of $H_2$ contains the most moles and therefore the greatest number of molecules. Students may incorrectly assume that all 1.0 g samples contain the same number of molecules.

---

4. B [2 marks]

Explanation:

• Moles of $CO_2 = 240 / 24000 = 0.010$ mol
• From the equation $MCO_3 + 2HCl \rightarrow MCl_2 + H_2O + CO_2$, moles of $MCO_3 = 0.010$ mol
• Molar mass of $MCO_3 = 2.00 / 0.010 = 200$ g mol$^{-1}$
• $M_r(M) + 12.0 + 3(16.0) = 200$
• $M_r(M) = 200 - 60.0 = 140$...

Wait — let me recalculate: $M_r(M) = 200 - 12.0 - 48.0 = 140$. This does not match any option. Let me re-examine.

Actually: $MCO_3$ molar mass = $M + 12 + 48 = M + 60$. So $M + 60 = 200$, giving $M = 140$. This doesn't match the options. Let me recheck the question setup.

Rechecking: If the answer is B (40.0), then molar mass of $MCO_3 = 40 + 60 = 100$ g mol$^{-1}$, and moles of $MCO_3 = 2.00/100 = 0.020$ mol, giving $CO_2$ volume = $0.020 \times 24000 = 480$ cm³. But the question states 240 cm³.

Let me re-examine: moles of $CO_2 = 240/24000 = 0.010$ mol. Molar mass of $MCO_3 = 2.00/0.010 = 200$. $M = 200 - 60 = 140$. None of the options match.

I need to fix the question. Let me adjust: if the mass is 1.00 g instead of 2.00 g, then molar mass of $MCO_3 = 1.00/0.010 = 100$, and $M = 40.0$ (Ca). Answer is B.

Corrected working for the question as stated (assuming the intended answer is B, M = 40.0):

• Moles of $CO_2 = 240 / 24000 = 0.010$ mol
• Moles of $MCO_3 = 0.010$ mol (1:1 ratio)
• Molar mass of $MCO_3 = 2.00 / 0.010 = 200$ g mol$^{-1}$

This gives $M = 140$, which is not among the options. The question should use 1.00 g of $MCO_3$ to give $M = 40.0$ (calcium). For marking purposes, the answer is B (40.0) assuming the question intended 1.00 g of metal carbonate.

Marking note: Award 2 marks for correct answer B with working. If a student identifies the inconsistency, award full marks for showing correct method.

---

5. C [2 marks]

Explanation:

• Molar mass of $H_2SO_4 = 2(1.0) + 32.1 + 4(16.0) = 98.1$ g mol$^{-1}$
• Moles of $H_2SO_4 = 4.90 / 98.1 = 0.04995 \approx 0.050$ mol
• Concentration = $0.050 / 0.250 = 0.20$ mol dm$^{-3}$

A common error is to forget to convert 250 cm³ to dm³ (divide by 1000), which would give 0.050/250 = 0.0002 mol dm$^{-3}$ — not among the options. Another error is using the mass directly: 4.90/250 = 0.0196 ≈ 0.020, which is not an option either.

---

Section B: Structured Questions

6. (a) Mole [2]

Answer: A mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 g of carbon-12. [2]

Teaching note: The mole is the SI unit for amount of substance. One mole of any substance contains the Avogadro constant ($6.02 \times 10^{23}$) of particles. Students should not define it merely as "a unit" — they must reference the number of particles or the carbon-12 standard.

(b) Avogadro constant [2]

Answer: The Avogadro constant is the number of particles (atoms, molecules, or ions) in one mole of a substance. Its value is $6.02 \times 10^{23}$ mol$^{-1}$. [2]

Teaching note: The Avogadro constant, $L$ (or $N_A$), bridges the macroscopic scale (grams, moles) with the atomic scale (number of particles). It is essential for converting between moles and number of particles.

---

7. (a) [2]

Working: $\text{Moles of Fe} = \frac{5.60}{55.8} = 0.100 \text{ mol} \quad [2]$

Award: 1 mark for correct formula/substitution, 1 mark for correct answer.

(b) [2]

Working: $\text{Number of atoms} = 0.100 \times 6.02 \times 10^{23} = 6.02 \times 10^{22} \quad [2]$

Award: 1 mark for using moles × Avogadro constant, 1 mark for correct answer. Follow through from (a) allowed.

---

8. (a) [2]

Working: $M_r(MgSO_4) = 24.3 + 32.1 + 4(16.0) = 120.4$ $\text{Moles of } MgSO_4 = \frac{6.00}{120.4} = 0.0498 \approx 0.050 \text{ mol} \quad [2]$

(b) [2]

Working: $\text{Mass of water removed} = 12.30 - 6.00 = 6.30 \text{ g}$ $M_r(H_2O) = 2(1.0) + 16.0 = 18.0$ $\text{Moles of } H_2O = \frac{6.30}{18.0} = 0.35 \text{ mol} \quad [2]$

(c) [2]

Working: $x = \frac{\text{moles of } H_2O}{\text{moles of } MgSO_4} = \frac{0.35}{0.050} = 7$

The formula is $MgSO_4 \cdot 7H_2O$. [2]

Common error: Students may divide moles of $MgSO_4$ by moles of $H_2O$ and get 0.14, or use mass ratio instead of mole ratio.

---

9. (a) [2]

Working:

Element	%	÷ $A_r$	Ratio
C	40.0	40.0/12.0 = 3.33	1
H	6.7	6.7/1.0 = 6.7	2
O	53.3	53.3/16.0 = 3.33	1

Dividing by smallest (3.33): C = 1, H = 2, O = 1. Empirical formula = $CH_2O$. [2]

(b) [2]

Working: $\text{Empirical formula mass} = 12.0 + 2(1.0) + 16.0 = 30.0 \text{ g mol}^{-1}$ $\text{Multiple} = \frac{60.0}{30.0} = 2$ $\text{Molecular formula} = (CH_2O)_2 = C_2H_4O_2 \quad [2]$

Teaching note: The molecular formula is a whole-number multiple of the empirical formula. Students must divide the molar mass by the empirical formula mass to find this multiple.

---

10. (a) [2]

Working: $\text{Moles of NaOH} = \frac{25.0}{1000} \times 0.100 = 0.00250 \text{ mol} \quad [2]$

(b) [2]

Working: From the equation: $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$ $\text{Mole ratio: } NaOH : H_2SO_4 = 2 : 1$ $\text{Moles of } H_2SO_4 = \frac{0.00250}{2} = 0.00125 \text{ mol} \quad [2]$

(c) [2]

Working: $\text{Concentration of } H_2SO_4 = \frac{0.00125}{15.0/1000} = \frac{0.00125}{0.0150} = 0.0833 \text{ mol dm}^{-3} \quad [2]$

Common error: Forgetting to convert cm³ to dm³. Also, using the wrong mole ratio (1:1 instead of 2:1) is a frequent mistake.

---

11. (a) [3]

Working: $M_r(N_2) = 2(14.0) = 28.0$ $\text{Moles of } N_2 = \frac{14.0}{28.0} = 0.50 \text{ mol}$

From the equation $N_2 + 3H_2 \rightarrow 2NH_3$: $\text{Mole ratio: } N_2 : NH_3 = 1 : 2$ $\text{Moles of } NH_3 = 2 \times 0.50 = 1.0 \text{ mol}$ $M_r(NH_3) = 14.0 + 3(1.0) = 17.0$ $\text{Mass of } NH_3 = 1.0 \times 17.0 = 17.0 \text{ g} \quad [3]$

Award: 1 mark for moles of $N_2$, 1 mark for moles of $NH_3$ (using ratio), 1 mark for mass.

(b) [2]

Working: $\text{Volume of } NH_3 = 1.0 \times 24.0 = 24.0 \text{ dm}^3 \quad [2]$

---

12. (a) [2]

Working: $M_r(KNO_3) = 39.1 + 14.0 + 3(16.0) = 101.1$ $\text{Moles of } KNO_3 = \frac{10.0}{101.1} = 0.0989 \approx 0.099 \text{ mol} \quad [2]$

(b) [2]

Working: From $2KNO_3 \rightarrow 2KNO_2 + O_2$: $\text{Mole ratio: } KNO_3 : O_2 = 2 : 1$ $\text{Moles of } O_2 = \frac{0.099}{2} = 0.0495 \text{ mol}$ $\text{Volume of } O_2 = 0.0495 \times 24.0 = 1.19 \text{ dm}^3 \quad [2]$

(c) [2]

Working: $\text{Percentage yield} = \frac{1.50}{1.19} \times 100\%$

Wait — the collected volume (1.50 dm³) is greater than the theoretical volume (1.19 dm³), which is impossible for a percentage yield. Let me fix this: the theoretical volume should be larger. If moles of $KNO_3 = 0.099$ mol, moles of $O_2 = 0.0495$ mol, theoretical volume = 1.19 dm³. The collected volume of 1.50 dm³ exceeds this.

I need to adjust the question. Let me change the collected volume to 0.95 dm³.

Corrected (c): $\text{Percentage yield} = \frac{0.95}{1.19} \times 100\% = 79.8\% \approx 80\% \quad [2]$

Award: 1 mark for correct formula, 1 mark for correct answer. Follow through allowed.

---

13. (a) [2]

Working: $\text{Moles of HCl} = \frac{50.0}{1000} \times 1.00 = 0.050 \text{ mol} \quad [2]$

(b) [2]

Working: From $MCO_3 + 2HCl \rightarrow MCl_2 + H_2O + CO_2$: $\text{Mole ratio: } MCO_3 : HCl = 1 : 2$

If all HCl reacted: moles of $MCO_3 = 0.050/2 = 0.025$ mol.

But we need to check if HCl is in excess. Moles of $MCO_3 = 2.50 / M_r(MCO_3)$. We don't know $M_r(MCO_3)$ yet, so we assume the reaction goes to completion with one reactant limiting.

Since the question states the carbonate was added to the acid, and asks us to identify M, we can work backwards. If moles of $MCO_3 = 0.025$ mol (HCl limiting), then $M_r(MCO_3) = 2.50/0.025 = 100$ g mol$^{-1}$, so $M_r(M) = 100 - 60 = 40.0$ (Ca).

Working: $\text{Moles of } MCO_3 = \frac{0.050}{2} = 0.025 \text{ mol} \quad [2]$

(c) [2]

Working: $M_r(MCO_3) = \frac{2.50}{0.025} = 100$ $M_r(M) = 100 - 12.0 - 3(16.0) = 100 - 60.0 = 40.0$

The metal M is calcium (Ca). [2]

---

14. (a) [2]

Working: $M_r(KIO_3) = 39.1 + 126.9 + 3(16.0) = 214.0$ $\text{Moles of } KIO_3 = \frac{0.317}{214.0} = 1.48 \times 10^{-3} \text{ mol} \quad [2]$

(b) [2]

Working: From the half-equations:

• $IO_3^- + 6H^+ + 5e^- \rightarrow \frac{1}{2}I_2 + 3H_2O$ — each $IO_3^-$ accepts 5 electrons
• $2S_2O_3^{2-} \rightarrow S_4O_6^{2-} + 2e^-$ — each $S_2O_3^{2-}$ donates 1 electron (2 thiosulfate ions donate 2 electrons)

To balance electrons: 1 mol $IO_3^-$ (accepts 5 e⁻) requires 5 mol $S_2O_3^{2-}$ (each donates 1 e⁻).

$\text{Mole ratio: } S_2O_3^{2-} : IO_3^- = 5 : 1 \quad [2]$

(c) [2]

Working: $\text{Moles of } S_2O_3^{2-} = 5 \times 1.48 \times 10^{-3} = 7.40 \times 10^{-3} \text{ mol}$ $\text{Concentration} = \frac{7.40 \times 10^{-3}}{25.0/1000} = \frac{7.40 \times 10^{-3}}{0.025} = 0.296 \text{ mol dm}^{-3} \quad [2]$

---

15. (a) [1]

Answer: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g) \quad [1]$

Note: $Na_2CO_3$ does not decompose under normal heating conditions (it has a very high decomposition temperature).

(b) [4]

Working: $M_r(Na_2CO_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0$ $M_r(CaCO_3) = 40.1 + 12.0 + 3(16.0) = 100.1$

Only $CaCO_3$ decomposes: $\text{Moles of } CaCO_3 = \frac{2.00}{100.1} = 0.01998 \approx 0.0200 \text{ mol}$

From the equation: moles of $CO_2 = 0.0200$ mol $\text{Volume of } CO_2 = 0.0200 \times 24.0 = 0.480 \text{ dm}^3 = 480 \text{ cm}^3 \quad [4]$

Award: 1 mark for moles of $CaCO_3$, 1 mark for identifying only $CaCO_3$ decomposes, 1 mark for moles of $CO_2$, 1 mark for volume.

(c) [3]

Working:

• Mass of $Na_2CO_3$ (unchanged) = 3.00 g
• Mass of $CaO$ produced:

$M_r(CaO) = 40.1 + 16.0 = 56.1$ $\text{Moles of CaO} = 0.0200 \text{ mol}$ $\text{Mass of CaO} = 0.0200 \times 56.1 = 1.12 \text{ g}$

$\text{Total residue} = 3.00 + 1.12 = 4.12 \text{ g} \quad [3]$

Award: 1 mark for mass of $Na_2CO_3$, 1 mark for mass of $CaO$, 1 mark for total.

---

Section C: Application and Data Interpretation

16. (a) [2]

Working: $\text{Percentage of Fe} = \frac{1.750}{2.500} \times 100\% = 70.0\% \quad [2]$

(b) [3]

Working: $M_r(Fe_2O_3) = 2(55.8) + 3(16.0) = 159.6$ $\text{Mass percentage of Fe in } Fe_2O_3 = \frac{2 \times 55.8}{159.6} \times 100\% = \frac{111.6}{159.6} \times 100\% = 69.9\%$

$\text{Percentage of } Fe_2O_3 \text{ in ore} = \frac{70.0}{69.9} \times 100\% = 100.1\% \approx 100\% \quad [3]$

Award: 1 mark for $M_r(Fe_2O_3)$, 1 mark for % Fe in pure $Fe_2O_3$, 1 mark for final answer.

(c) [1]

Answer: The ore may contain other iron-containing impurities (such as $Fe_3O_4$ or other iron compounds), or the extraction may not have been 100% efficient, or the ore may contain moisture or other non-iron components. [1]

Accept any reasonable suggestion.

---

17. (a) [1]

Answer: Maximum volume of $CO_2$ = 240 cm³ (read from the plateau of the graph). [1]

(b) [2]

Working: $\text{Moles of } CO_2 = \frac{240}{24000} = 0.010 \text{ mol} \quad [2]$

(c) [2]

Working: From $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$: $\text{Mole ratio: } HCl : CO_2 = 2 : 1$ $\text{Moles of HCl} = 2 \times 0.010 = 0.020 \text{ mol} \quad [2]$

(d) [2]

Answer: As the reaction proceeds, the concentration (and therefore the number of particles per unit volume) of $HCl$ decreases. This means there are fewer $HCl$ particles available to collide with the $CaCO_3$ surface per unit time. The frequency of effective collisions decreases, so the rate of reaction decreases. [2]

Award: 1 mark for mentioning decreasing concentration of reactant, 1 mark for linking to collision frequency/effective collisions.

---

18. (a) [1]

Answer: The rough titration is a preliminary trial to estimate the approximate volume needed. It is less accurate and is not used in the calculation of the average titre to improve reliability. [1]

(b) [2]

Working: Titration 1: 23.80 − 0.00 = 23.80 cm³ Titration 2: 23.70 − 0.00 = 23.70 cm³ Titration 3: 23.90 − 0.20 = 23.70 cm³

Average titre = $\frac{23.80 + 23.70 + 23.70}{3} = \frac{71.20}{3} = 23.73$ cm³ ≈ 23.7 cm³ [2]

Note: Titrations 2 and 3 are concordant (within 0.10 cm³). Titration 1 is also within 0.10 cm³ of the others, so all three are used.

(c) [3]

Working: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$ $\text{Mole ratio} = 1 : 1$

$\text{Moles of NaOH} = \frac{23.7}{1000} \times 0.100 = 0.00237 \text{ mol}$ $\text{Moles of } CH_3COOH = 0.00237 \text{ mol}$ $\text{Concentration of } CH_3COOH = \frac{0.00237}{25.0/1000} = \frac{0.00237}{0.025} = 0.0948 \text{ mol dm}^{-3} \quad [3]$

Award: 1 mark for moles of NaOH, 1 mark for mole ratio application, 1 mark for concentration.

---

19. (a) [1]

Answer: The 30.0 cm³ of gas remaining after passing through sodium hydroxide is unreacted (excess) oxygen. Sodium hydroxide absorbs $CO_2$, so the remaining gas must be oxygen that did not react. [1]

(b) [2]

Working: $\text{Volume of } CO_2 = 110.0 - 30.0 = 80.0 \text{ cm}^3 \quad [2]$

(c) [2]

Working: Initial volume of $O_2$ = 150.0 cm³ Volume of excess $O_2$ = 30.0 cm³ $\text{Volume of } O_2 \text{ used} = 150.0 - 30.0 = 120.0 \text{ cm}^3 \quad [2]$

(d) [3]

Working: The general combustion equation is: $C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$

At constant temperature and pressure, volume ratios = mole ratios.

Volume of hydrocarbon = 20.0 cm³ Volume of $CO_2$ produced = 80.0 cm³ Volume of $O_2$ used = 120.0 cm³

From the equation: $\frac{V_{CO_2}}{V_{hydrocarbon}} = \frac{x}{1} = \frac{80.0}{20.0} = 4 \Rightarrow x = 4$ $\frac{V_{O_2}}{V_{hydrocarbon}} = \frac{x + y/4}{1} = \frac{120.0}{20.0} = 6$ $4 + \frac{y}{4} = 6 \Rightarrow \frac{y}{4} = 2 \Rightarrow y = 8$

Molecular formula = $C_4H_8$ [3]

Award: 1 mark for finding x = 4, 1 mark for setting up equation for y, 1 mark for y = 8 and molecular formula.

---

20. (a) [2]

Working: $\text{Moles of NaOH} = \frac{25.0}{1000} \times 0.200 = 0.00500 \text{ mol} \quad [2]$

(b) [1]

Working: From $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$: $\text{Mole ratio: } H_2SO_4 : NaOH = 1 : 2$ $\text{Moles of excess } H_2SO_4 = \frac{0.00500}{2} = 0.00250 \text{ mol} \quad [1]$

(c) [2]

Working: $\text{Initial moles of } H_2SO_4 = \frac{50.0}{1000} \times 0.500 = 0.0250 \text{ mol}$ $\text{Moles of } H_2SO_4 \text{ that reacted with } NH_3 = 0.0250 - 0.00250 = 0.0225 \text{ mol} \quad [2]$

(d) [3]

Working: From $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$: $\text{Mole ratio: } NH_3 : H_2SO_4 = 2 : 1$ $\text{Moles of } NH_3 = 2 \times 0.0225 = 0.0450 \text{ mol}$

From $(NH_4)_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2NH_3 + 2H_2O$: $\text{Mole ratio: } (NH_4)_2SO_4 : NH_3 = 1 : 2$ $\text{Moles of } (NH_4)_2SO_4 = \frac{0.0450}{2} = 0.0225 \text{ mol}$

This is the amount in 25.0 cm³ of the solution. The total solution is 250 cm³, so: $\text{Total moles of } (NH_4)_2SO_4 = 0.0225 \times \frac{250}{25.0} = 0.225 \text{ mol}$

$M_r((NH_4)_2SO_4) = 2(14.0) + 8(1.0) + 32.1 + 4(16.0) = 132.1$ $\text{Mass of } (NH_4)_2SO_4 = 0.225 \times 132.1 = 29.7 \text{ g}$

Wait — this exceeds the 5.00 g sample mass. Let me recheck.

The moles of $(NH_4)_2SO_4$ in 25.0 cm³ of solution = 0.0225 mol. Total moles in 250 cm³ = $0.0225 \times 10 = 0.225$ mol. Mass = $0.225 \times 132.1 = 29.7$ g. This is impossible since the sample is only 5.00 g.

I need to fix the numbers. Let me adjust: if the NaOH concentration for neutralisation is 0.200 mol dm⁻³ and volume is 25.0 cm³, moles of NaOH = 0.00500 mol, moles of excess $H_2SO_4$ = 0.00250 mol, moles of $H_2SO_4$ that reacted with $NH_3$ = 0.0250 − 0.00250 = 0.0225 mol. This is correct.

The issue is the scaling. The 0.0225 mol of $H_2SO_4$ reacted with $NH_3$ from 25.0 cm³ of the fertiliser solution. The total solution is 250 cm³, so total moles of $(NH_4)_2SO_4 = 0.0225/2 \times 10 = 0.1125$ mol. Mass = $0.1125 \times 132.1 = 14.86$ g. Still too high.

Let me adjust the initial $H_2SO_4$ volume to 25.0 cm³ instead of 50.0 cm³:

• Initial moles of $H_2SO_4 = 25.0/1000 \times 0.500 = 0.0125$ mol
• Excess $H_2SO_4 = 0.00250$ mol
• $H_2SO_4$ that reacted with $NH_3 = 0.0125 - 0.00250 = 0.0100$ mol
• Moles of $NH_3 = 2 \times 0.0100 = 0.0200$ mol (from 25.0 cm³)
• Moles of $(NH_4)_2SO_4$ in 25.0 cm³ = $0.0200/2 = 0.0100$ mol
• Total moles in 250 cm³ = $0.0100 \times 10 = 0.100$ mol
• Mass of $(NH_4)_2SO_4 = 0.100 \times 132.1 = 13.21$ g. Still too high for a 5.00 g sample.

Let me adjust the NaOH volume for neutralisation to 20.0 cm³:

• Moles of NaOH = $20.0/1000 \times 0.200 = 0.00400$ mol
• Excess $H_2SO_4 = 0.00400/2 = 0.00200$ mol
• $H_2SO_4$ reacted with $NH_3 = 0.0125 - 0.00200 = 0.0105$ mol
• Moles of $NH_3 = 2 \times 0.0105 = 0.0210$ mol
• Moles of $(NH_4)_2SO_4$ in 25.0 cm³ = $0.0210/2 = 0.0105$ mol
• Total in 250 cm³ = $0.0105 \times 10 = 0.105$ mol
• Mass = $0.105 \times 132.1 = 13.87$ g. Still too high.

The fundamental issue is that the numbers need to give a mass less than 5.00 g. Let me redesign this part:

If the mass of $(NH_4)_2SO_4$ in 5.00 g of fertiliser is to be, say, 3.96 g (79.2% purity):

• Moles of $(NH_4)_2SO_4 = 3.96/132.1 = 0.0300$ mol (in 250 cm³)
• Moles in 25.0 cm³ = $0.0300/10 = 0.00300$ mol
• Moles of $NH_3 = 2 \times 0.00300 = 0.00600$ mol
• Moles of $H_2SO_4$ that reacted with $NH_3 = 0.00600/2 = 0.00300$ mol
• Initial moles of $H_2SO_4 = 50.0/1000 \times 0.500 = 0.0250$ mol
• Excess $H_2SO_4 = 0.0250 - 0.00300 = 0.0220$ mol
• Moles of NaOH needed = $2 \times 0.0220 = 0.0440$ mol
• Volume of NaOH = $0.0440/0.200 = 0.220$ dm³ = 220 cm³

This is too large. Let me use a different approach — reduce the initial $H_2SO_4$:

If initial $H_2SO_4$ is 25.0 cm³ of 0.200 mol dm⁻³:

• Initial moles = $25.0/1000 \times 0.200 = 0.00500$ mol
• Excess $H_2SO_4 = 0.00200$ mol (from 20.0 cm³ of 0.200 M NaOH)
• $H_2SO_4$ reacted with $NH_3 = 0.00500 - 0.00200 = 0.00300$ mol
• Moles of $NH_3 = 0.00600$ mol
• Moles of $(NH_4)_2SO_4$ in 25.0 cm³ = 0.00300 mol
• Total in 250 cm³ = 0.0300 mol
• Mass = $0.0300 \times 132.1 = 3.96$ g ✓ (less than 5.00 g)
• Percentage = $3.96/5.00 \times 100\% = 79.2\%$ ✓

I need to adjust the question parameters. Let me revise Q20:

Revised Q20 parameters:

• 50.0 cm³ of 0.200 mol dm⁻³ $H_2SO_4$ (not 0.500)
• 20.0 cm³ of 0.200 mol dm⁻³ NaOH for neutralisation

Let me redo the answer key with corrected values:

(a) [2] $\text{Moles of NaOH} = \frac{20.0}{1000} \times 0.200 = 0.00400 \text{ mol} \quad [2]$

(b) [1] $\text{Moles of excess } H_2SO_4 = \frac{0.00400}{2} = 0.00200 \text{ mol} \quad [1]$

(c) [2] $\text{Initial moles of } H_2SO_4 = \frac{50.0}{1000} \times 0.200 = 0.0100 \text{ mol}$ $\text{Moles of } H_2SO_4 \text{ reacted with } NH_3 = 0.0100 - 0.00200 = 0.00800 \text{ mol} \quad [2]$

(d) [3] $\text{Moles of } NH_3 = 2 \times 0.00800 = 0.0160 \text{ mol}$ $\text{Moles of } (NH_4)_2SO_4 \text{ in 25.0 cm}^3 = \frac{0.0160}{2} = 0.00800 \text{ mol}$ $\text{Total moles in 250 cm}^3 = 0.00800 \times \frac{250}{25.0} = 0.0800 \text{ mol}$ $\text{Mass of } (NH_4)_2SO_4 = 0.0800 \times 132.1 = 10.57 \text{ g}$

Still too high. The problem is the scaling factor of 10. Let me use a 500 cm³ solution instead, or reduce the sample mass, or use different concentrations.

Let me try: 5.00 g fertiliser dissolved in 500 cm³, 25.0 cm³ used:

• Total moles in 500 cm³ = moles in 25.0 cm³ × 20

Or: use 10.0 cm³ of the solution for the reaction:

• Scaling factor = 250/10 = 25

Let me try yet another approach. Use 0.100 mol dm⁻³ $H_2SO_4$ (50.0 cm³) and 0.100 mol dm⁻³ NaOH (20.0 cm³):

• Initial moles $H_2SO_4 = 50.0/1000 \times 0.100 = 0.00500$ mol
• Moles NaOH = $20.0/1000 \times 0.100 = 0.00200$ mol
• Excess $H_2SO_4 = 0.00200/2 = 0.00100$ mol
• $H_2SO_4$ reacted with $NH_3 = 0.00500 - 0.00100 = 0.00400$ mol
• Moles $NH_3 = 2 \times 0.00400 = 0.00800$ mol
• Moles $(NH_4)_2SO_4$ in 25.0 cm³ = 0.00400 mol
• Total in 250 cm³ = $0.00400 \times 10 = 0.0400$ mol
• Mass = $0.0400 \times 132.1 = 5.28$ g — still slightly over 5.00 g.

Let me use 0.080 mol dm⁻³ $H_2SO_4$:

• Initial moles = $50.0/1000 \times 0.080 = 0.00400$ mol
• Excess $H_2SO_4 = 0.00100$ mol (from 20.0 cm³ of 0.100 M NaOH)
• $H_2SO_4$ reacted = $0.00400 - 0.00100 = 0.00300$ mol
• Moles $NH_3 = 0.00600$ mol
• Moles $(NH_4)_2SO_4$ in 25.0 cm³ = 0.00300 mol
• Total in 250 cm³ = 0.0300 mol
• Mass = $0.0300 \times 132.1 = 3.96$ g ✓
• Percentage = $3.96/5.00 \times 100\% = 79.2\%$ ✓

This works. But I need to update the question. Let me use these revised parameters in the answer key and note the correction.

For the answer key, I will use the corrected parameters:

• 50.0 cm³ of 0.080 mol dm⁻³ $H_2SO_4$
• 20.0 cm³ of 0.100 mol dm⁻³ NaOH for neutralisation

(a) [2] $\text{Moles of NaOH} = \frac{20.0}{1000} \times 0.100 = 0.00200 \text{ mol} \quad [2]$

(b) [1] $\text{Moles of excess } H_2SO_4 = \frac{0.00200}{2} = 0.00100 \text{ mol} \quad [1]$

(c) [2] $\text{Initial moles of } H_2SO_4 = \frac{50.0}{1000} \times 0.080 = 0.00400 \text{ mol}$ $\text{Moles of } H_2SO_4 \text{ reacted with } NH_3 = 0.00400 - 0.00100 = 0.00300 \text{ mol} \quad [2]$

(d) [3] $\text{Moles of } NH_3 = 2 \times 0.00300 = 0.00600 \text{ mol}$ $\text{Moles of } (NH_4)_2SO_4 \text{ in 25.0 cm}^3 = \frac{0.00600}{2} = 0.00300 \text{ mol}$ $\text{Total moles in 250 cm}^3 = 0.00300 \times \frac{250}{25.0} = 0.0300 \text{ mol}$ $\text{Mass of } (NH_4)_2SO_4 = 0.0300 \times 132.1 = 3.96 \text{ g} \quad [3]$

Award: 1 mark for moles of $NH_3$, 1 mark for scaling to total solution, 1 mark for mass.

(e) [1] $\text{Percentage by mass} = \frac{3.96}{5.00} \times 100\% = 79.2\% \quad [1]$

---

Mark Summary

Section	Questions	Marks
A: Multiple Choice	1–5	10
B: Structured Questions	6–15	30
C: Application & Data Interpretation	16–20	20
Total	20 questions	60