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Secondary 3 Chemistry Semestral Assessment 2 (End of Year) Paper 2
Free Sec 3 Chemistry SA2 Paper 2, Nemo3 Exam version, with questions, answers, and O Level-style practice for Singapore students.
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Questions
TuitionGoWhere Practice Paper - Chemistry Secondary 3
TuitionGoWhere Secondary School (AI)
Subject: Chemistry
Level: Secondary 3 (Express/G3)
Paper: SA2 Version 2
Duration: 1 hour 30 minutes
Total Marks: 60
Name: ________________________
Class: ________________________
Date: ________________________
INSTRUCTIONS TO CANDIDATES
- Write your name, class, and date in the spaces provided above.
- Answer all questions in the spaces provided on the question paper.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- The total number of marks for this paper is 60.
- You are advised to spend approximately 45 minutes on Section A and 45 minutes on Section B.
- A copy of the Periodic Table is printed on page 2.
- For calculations, show all working clearly. Marks may be awarded for correct method even if the final answer is incorrect.
- Use appropriate significant figures and include units in your final answers where applicable.
Periodic Table (Simplified)
| Group → | 1 | 2 | 13 | 14 | 15 | 16 | 17 | 0 |
|---|---|---|---|---|---|---|---|---|
| Period 1 | H | He | ||||||
| Period 2 | Li | Be | B | C | N | O | F | Ne |
| Period 3 | Na | Mg | Al | Si | P | S | Cl | Ar |
| Period 4 | K | Ca | ... | ... | ... | ... | Br | Kr |
Relative atomic masses: H=1, He=4, Li=7, Be=9, B=11, C=12, N=14, O=16, F=19, Ne=20, Na=23, Mg=24, Al=27, Si=28, P=31, S=32, Cl=35.5, Ar=40, K=39, Ca=40, Br=80, Kr=84
Section A: Structured Questions [30 marks]
Answer all questions in the spaces provided.
Question 1
A student tests the pH of four different soil samples using universal indicator. The results are shown in the table below.
| Soil Sample | Colour with Universal Indicator | pH Value |
|---|---|---|
| A | Yellow | 5.5 |
| B | Green | 7.0 |
| C | Blue | 8.5 |
| D | Red | 4.0 |
(a) Which soil sample is the most acidic? [1]
(b) A farmer wants to grow a crop that grows best in slightly alkaline soil (pH 7.5–8.0). Which soil sample is most suitable without any treatment? [1]
(c) The farmer decides to treat Soil Sample D by adding solid calcium hydroxide, Ca(OH)₂. Write a balanced chemical equation, including state symbols, for the reaction that occurs between calcium hydroxide and the acidic components in the soil. [2]
(d) Explain why solid calcium carbonate, CaCO₃, would be less effective than calcium hydroxide at neutralising the same acidic soil. [2]
Question 2
The diagram below shows the apparatus used to prepare a soluble salt by titration.
Image pending generation: experimental_setup for Q2.
(a) Name the piece of apparatus labelled as the burette. [1]
(b) State the colour change observed in the conical flask at the endpoint of the titration. [1]
(c) Calculate the number of moles of HCl in the conical flask. [1]
(d) Using your answer to (c), determine the volume of NaOH required to reach the endpoint. Show your working. [2]
(e) After the titration, the student evaporates the neutral solution to obtain solid sodium chloride. Describe how the student should obtain pure, dry crystals of sodium chloride from the neutral solution. [3]
Question 3
Ammonium sulfate, (NH₄)₂SO₄, is a common fertiliser. It can be prepared by reacting aqueous ammonia with sulfuric acid.
(a) Write the balanced chemical equation, including state symbols, for the reaction between aqueous ammonia and sulfuric acid. [2]
(b) A student adds 50.0 cm³ of 1.00 mol/dm³ sulfuric acid to excess aqueous ammonia. Calculate the maximum mass of ammonium sulfate that can be obtained. (Mᵣ of (NH₄)₂SO₄ = 132) [3]
(c) The student obtains 5.80 g of ammonium sulfate crystals. Calculate the percentage yield. [2]
(d) Suggest one reason why the actual yield is less than the theoretical yield. [1]
Question 4
The table below shows the results of adding different reagents to three unknown solutions, X, Y, and Z.
| Test | Solution X | Solution Y | Solution Z |
|---|---|---|---|
| Add aqueous NaOH (excess) | White precipitate, soluble in excess | Light blue precipitate, insoluble in excess | No visible change |
| Add aqueous NH₃ (excess) | White precipitate, insoluble in excess | Light blue precipitate, soluble in excess (deep blue) | No visible change |
| Add dilute HCl, then BaCl₂ (aq) | No visible change | No visible change | White precipitate |
| Add dilute HNO₃, then AgNO₃ (aq) | White precipitate | No visible change | No visible change |
(a) Identify the cation present in Solution X. [1]
(b) Identify the cation present in Solution Y. [1]
(c) Identify the anion present in Solution Z. [1]
(d) Write the ionic equation, including state symbols, for the reaction that occurs when aqueous NaOH is added to Solution X until in excess. [2]
(e) Solution Y contains a transition metal cation. Explain why the precipitate formed with aqueous NH₃ dissolves in excess to form a deep blue solution. [2]
Question 5
A student investigates the rate of reaction between calcium carbonate chips and dilute hydrochloric acid. The apparatus used is shown below.
Image pending generation: experimental_setup for Q5.
The student measures the volume of carbon dioxide gas produced every 30 seconds. The results are plotted on the graph below.
Image pending generation: graph for Q5.
(a) Write the balanced chemical equation, including state symbols, for the reaction between calcium carbonate and hydrochloric acid. [2]
(b) Using the graph, determine the average rate of reaction in the first 30 seconds for the original experiment (1.0 mol/dm³ HCl). Give your answer in cm³/s. [2]
(c) Explain, in terms of collision theory, why Curve A (2.0 mol/dm³ HCl) is steeper than the original curve. [2]
(d) The student repeats the experiment using the same mass of calcium carbonate but in powdered form instead of chips. Sketch on the graph (or describe) how the curve would differ from the original. Explain your answer. [3]
Section B: Free Response / Data-Based Questions [30 marks]
Answer all questions in the spaces provided.
Question 6
Acid rain is a major environmental problem caused by emissions of sulfur dioxide and nitrogen oxides from industrial processes and vehicles. These gases dissolve in atmospheric water to form acids.
(a) Write a balanced chemical equation for the formation of sulfuric acid from sulfur dioxide in the atmosphere. Include state symbols. [2]
(b) Limestone (calcium carbonate) statues and buildings are damaged by acid rain. Write the ionic equation for the reaction between calcium carbonate and sulfuric acid. [2]
(c) Lakes in areas with granite bedrock are more severely affected by acid rain than lakes in areas with limestone bedrock. Explain this observation. [3]
(d) One method to treat acidified lakes is to add powdered limestone. A lake has a volume of 2.5 × 10⁶ m³ and a pH of 4.5. Calculate the minimum mass of calcium carbonate (in tonnes) required to neutralise the acid in the lake, assuming the acid is entirely sulfuric acid and the target pH is 7.0. (1 m³ = 1000 dm³; Mᵣ of CaCO₃ = 100) [4]
Question 7
A student is given three unlabelled bottles containing colourless solutions: hydrochloric acid (HCl), sodium hydroxide (NaOH), and sodium chloride (NaCl). The student has only red and blue litmus paper, and a piece of magnesium ribbon.
(a) Describe a sequence of tests using only these materials to identify each solution. [4]
(b) For the solution identified as hydrochloric acid, write the ionic equation for its reaction with magnesium. [1]
(c) The student then tests the electrical conductivity of each solution. Predict and explain the relative conductivity of the three solutions. [3]
Question 8
The preparation of insoluble salts requires a precipitation reaction. A student wishes to prepare a pure, dry sample of lead(II) sulfate, PbSO₄.
(a) Name two suitable soluble reactants that can be used to prepare lead(II) sulfate by precipitation. [1]
(b) Write the ionic equation, including state symbols, for the precipitation reaction. [2]
(c) Describe the complete procedure to obtain a pure, dry sample of lead(II) sulfate from the reaction mixture, starting from the two aqueous reactants. [4]
(d) Lead(II) sulfate is a white solid. When heated strongly, it decomposes to form lead(II) oxide, sulfur dioxide, and oxygen. Write a balanced chemical equation for this thermal decomposition. [2]
Question 9
The diagram below shows the pH changes during the titration of a weak acid (ethanoic acid, CH₃COOH) with a strong base (sodium hydroxide, NaOH).
Image pending generation: graph for Q9.
(a) Explain why the pH at the equivalence point is greater than 7. [2]
(b) At the half-equivalence point (12.5 cm³ of NaOH added), the pH equals the pKₐ of ethanoic acid. State the pKₐ value from the graph. [1]
(c) A student suggests using methyl orange (pH range 3.1–4.4) as the indicator for this titration. Explain why this would be unsuitable. [2]
(d) The student titrates 25.0 cm³ of 0.100 mol/dm³ ethanoic acid with 0.100 mol/dm³ NaOH. Calculate the pH of the solution after 12.5 cm³ of NaOH has been added. (Kₐ of CH₃COOH = 1.74 × 10⁻⁵ mol/dm³) [3]
Question 10
Fertilisers often contain ammonium compounds. A student analyses a sample of ammonium nitrate fertiliser, NH₄NO₃, to determine its nitrogen content.
(a) Calculate the percentage by mass of nitrogen in ammonium nitrate. (Mᵣ of NH₄NO₃ = 80) [2]
(b) The student heats a 2.00 g sample of the fertiliser with aqueous sodium hydroxide and drives off the ammonia gas produced into a known volume of standard hydrochloric acid. The reaction is: NH₄NO₃ + NaOH → NaNO₃ + NH₃ + H₂O
The unreacted HCl is then titrated with 0.100 mol/dm³ NaOH. 25.0 cm³ of 0.500 mol/dm³ HCl was used initially, and 18.5 cm³ of 0.100 mol/dm³ NaOH was required to neutralise the excess HCl.
(i) Calculate the number of moles of HCl that reacted with the ammonia gas. [2]
(ii) Calculate the mass of nitrogen in the fertiliser sample. [2]
(iii) Calculate the percentage purity of the ammonium nitrate in the fertiliser sample. [2]
END OF PAPER
Total Marks: 60
Answers
TuitionGoWhere Practice Paper - Chemistry Secondary 3 SA2 Version 2 - Answer Key
Total Marks: 60
Section A: Structured Questions [30 marks]
Question 1
(a) Soil Sample D [1]
Reasoning: Sample D has the lowest pH (4.0) and shows red with universal indicator, indicating it is the most acidic.
(b) Soil Sample C [1]
Reasoning: Sample C has pH 8.5 (blue with universal indicator), which falls within the slightly alkaline range (7.5–8.0) required by the crop. Sample B is neutral (pH 7.0).
(c) Ca(OH)₂(s) + 2H⁺(aq) → Ca²⁺(aq) + 2H₂O(l) [2]
Mark breakdown: Correct reactants and products [1], correct balancing and state symbols [1]
Alternative acceptable: Ca(OH)₂(s) + 2H₃O⁺(aq) → Ca²⁺(aq) + 3H₂O(l)
Common mistake: Writing H₂SO₄ or "acid" instead of H⁺/H₃O⁺; omitting state symbols.
(d) Calcium carbonate is a weaker base / less soluble than calcium hydroxide. [1] It reacts more slowly and incompletely with acid because it is sparingly soluble in water, so fewer carbonate ions are available to neutralise H⁺. [1] Calcium hydroxide is more soluble (stronger alkali), providing a higher concentration of OH⁻ ions for faster, more complete neutralisation. [1]
Key concept: Solubility affects rate and extent of neutralisation. Ca(OH)₂ is a strong base (fully dissociated); CaCO₃ is a weak base (sparingly soluble).
Question 2
(a) Burette [1]
(b) Colourless to pink [1]
Phenolphthalein is colourless in acid/neutral, pink in alkali. At endpoint, slight excess NaOH turns it pink.
(c) Moles of HCl = concentration × volume (in dm³) = 0.100 mol/dm³ × 0.0250 dm³ = 0.00250 mol [1]
(d) Reaction: HCl + NaOH → NaCl + H₂O (1:1 mole ratio)
Moles of NaOH needed = moles of HCl = 0.00250 mol [1]
Volume of NaOH = moles ÷ concentration = 0.00250 mol ÷ 0.100 mol/dm³ = 0.0250 dm³ = 25.0 cm³ [1]
Matches burette reading (0.00 to 25.00 cm³).
(e) Procedure for pure, dry NaCl crystals:
- Pour the neutral solution into an evaporating dish. [1]
- Heat gently (over a water bath or low flame) to evaporate most of the water until a saturated solution forms / crystals start to appear at the edge. [1]
- Allow to cool slowly to room temperature for crystallisation. [1]
- Filter the crystals, wash with a small amount of cold distilled water, and dry between filter papers / in a low-temperature oven. [1]
Total 3 marks. Key points: gentle evaporation to saturation, cooling for crystallisation, filtration, washing, drying. Common mistake: "heat to dryness" — this causes spattering and impurity inclusion.
Question 3
(a) 2NH₃(aq) + H₂SO₄(aq) → (NH₄)₂SO₄(aq) [2]
Mark breakdown: Correct formulae [1], correct balancing and state symbols [1]
Note: NH₃(aq) not NH₄OH; H₂SO₄ and product are aqueous.
(b) Moles of H₂SO₄ = 1.00 mol/dm³ × 0.0500 dm³ = 0.0500 mol [1]
Mole ratio H₂SO₄ : (NH₄)₂SO₄ = 1 : 1
Moles of (NH₄)₂SO₄ = 0.0500 mol [1]
Mass = moles × Mᵣ = 0.0500 × 132 = 6.60 g [1]
(c) Percentage yield = (actual yield ÷ theoretical yield) × 100% = (5.80 ÷ 6.60) × 100% = 87.9% [2]
Mark breakdown: Correct formula/substitution [1], correct calculation [1]
(d) Any one valid reason:
- Some product lost during filtration/washing/transfer [1]
- Reaction did not go to completion (reversible/equilibrium) [1]
- Impurities in reactants / side reactions [1]
- Incomplete crystallisation (some product remains in solution) [1]
Question 4
(a) Aluminium ion, Al³⁺ [1]
Reasoning: White ppt with NaOH soluble in excess → Al³⁺, Zn²⁺, or Pb²⁺. With NH₃, white ppt insoluble in excess → Al³⁺ (Zn²⁺ and Pb²⁺ dissolve in excess NH₃).
(b) Copper(II) ion, Cu²⁺ [1]
Reasoning: Light blue ppt with NaOH insoluble in excess → Cu²⁺. With NH₃, light blue ppt soluble in excess forming deep blue [Cu(NH₃)₄]²⁺ solution → confirms Cu²⁺.
(c) Sulfate ion, SO₄²⁻ [1]
Reasoning: White ppt with BaCl₂ after acidification with HCl → SO₄²⁻ (BaSO₄ insoluble). No reaction with AgNO₃ after HNO₃ → not halide.
(d) Al³⁺(aq) + 3OH⁻(aq) → Al(OH)₃(s) [1]
Al(OH)₃(s) + OH⁻(aq) → [Al(OH)₄]⁻(aq) [1]
Or combined: Al³⁺(aq) + 4OH⁻(aq) → [Al(OH)₄]⁻(aq)
Mark breakdown: First equation (precipitation) [1], second equation (dissolving in excess) [1]. State symbols required.
(e) Cu²⁺ forms a light blue precipitate Cu(OH)₂ with NH₃. In excess NH₃, the precipitate dissolves due to ligand exchange: water ligands are replaced by ammonia ligands, forming the tetraamminecopper(II) complex ion [Cu(NH₃)₄(H₂O)₂]²⁺ (often written as [Cu(NH₃)₄]²⁺), which is deep blue and soluble. [2]
Key points: Ligand substitution / complex ion formation; ammonia acts as a ligand; colour change due to d-orbital splitting in the complex.
Question 5
(a) CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) [2]
Mark breakdown: Correct formulae [1], correct balancing and state symbols [1]
(b) From graph (original curve): Volume at 30 s ≈ 35 cm³ (read from graph), Volume at 0 s = 0 cm³
Average rate = (35 − 0) cm³ ÷ 30 s = 1.17 cm³/s (accept 1.1–1.3 cm³/s depending on graph reading) [2]
Mark breakdown: Correct reading from graph [1], correct calculation with units [1]
(c) Higher concentration (2.0 mol/dm³) means more HCl particles per unit volume. [1] This increases the frequency of effective collisions between H⁺ ions and CaCO₃ surface per unit time, increasing the rate of reaction. [1]
Collision theory: Rate ∝ collision frequency ∝ concentration of reactants.
(d) Powdered CaCO₃ has a larger total surface area than chips of the same mass. [1] More surface exposed to acid → more frequent collisions → faster initial rate (steeper initial gradient). [1] Same final volume of CO₂ (same limiting reagent, CaCO₃), so curve levels off at same volume but reaches it sooner. [1]
Sketch description: Curve starts steeper than original, reaches ~90 cm³ earlier (e.g., ~120 s), same plateau.
Section B: Free Response / Data-Based Questions [30 marks]
Question 6
(a) 2SO₂(g) + O₂(g) + 2H₂O(l) → 2H₂SO₄(aq) [2]
Or: SO₂(g) + ½O₂(g) + H₂O(l) → H₂SO₄(aq)
Mark breakdown: Correct reactants/products [1], correct balancing and state symbols [1]
Note: In atmosphere, SO₂ oxidised to SO₃ then dissolves; simplified as above.
(b) CO₃²⁻(s) + 2H⁺(aq) → CO₂(g) + H₂O(l) [2]
Mark breakdown: Correct ions [1], correct products and balancing [1]
Ca²⁺ is spectator; solid CaCO₃ provides CO₃²⁻ at surface.
(c) Granite is composed of silicate minerals (e.g., feldspar, quartz) which are insoluble and do not react with acid. [1] No neutralisation occurs, so acid accumulates in the lake. [1] Limestone (CaCO₃) reacts with acid rain (neutralisation), buffering the lake pH near 7. [1]
Key concept: Buffering capacity of carbonate bedrock vs. inert silicate bedrock.
(d) Target pH 7.0 → [H⁺] = 10⁻⁷ mol/dm³ (negligible compared to initial).
Initial pH 4.5 → [H⁺] = 10⁻⁴·⁵ = 3.16 × 10⁻⁵ mol/dm³
Volume of lake = 2.5 × 10⁶ m³ = 2.5 × 10⁹ dm³
Moles of H⁺ = (3.16 × 10⁻⁵ mol/dm³) × (2.5 × 10⁹ dm³) = 7.90 × 10⁴ mol [1]
H₂SO₄ provides 2 H⁺ per mole → moles of H₂SO₄ = 3.95 × 10⁴ mol
Reaction: CaCO₃ + H₂SO₄ → CaSO₄ + CO₂ + H₂O (1:1 mole ratio CaCO₃ : H₂SO₄)
Moles of CaCO₃ needed = 3.95 × 10⁴ mol [1]
Mass = moles × Mᵣ = 3.95 × 10⁴ × 100 = 3.95 × 10⁶ g = 3.95 tonnes [1]
Alternative: If treating H⁺ directly: CaCO₃ + 2H⁺ → Ca²⁺ + CO₂ + H₂O, moles CaCO₃ = ½ × moles H⁺ = 3.95 × 10⁴ mol → same answer.
Mark breakdown: [H⁺] calculation [1], lake volume conversion [1], mole ratio [1], final mass in tonnes [1]
Question 7
(a) Test sequence:
- Test each solution with blue litmus paper. The solution that turns blue litmus red is HCl (acidic). [1]
- Test the remaining two solutions with red litmus paper. The solution that turns red litmus blue is NaOH (alkaline). [1]
- The solution that causes no colour change to either litmus paper is NaCl (neutral). [1]
- Confirm HCl by adding magnesium ribbon to the identified acid: effervescence (H₂ gas) observed. No reaction with NaOH or NaCl. [1]
Total 4 marks. Logical sequence using only provided materials.
(b) Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g) [1]
Cl⁻ is spectator ion.
(c) HCl: High conductivity — strong acid, fully dissociated into H⁺ and Cl⁻ ions (high mobile ion concentration). [1]
NaOH: High conductivity — strong base, fully dissociated into Na⁺ and OH⁻ ions (high mobile ion concentration). [1]
NaCl: Moderate conductivity — soluble salt, fully dissociated into Na⁺ and Cl⁻, but typically lower concentration in this context (or same concentration but only 2 ions vs. H⁺/OH⁻ which have higher mobility). [1]
Relative order: HCl ≈ NaOH > NaCl (if same molarity, H⁺ and OH⁻ have anomalously high mobility due to proton hopping/Grotthuss mechanism). Explanation must mention: complete dissociation → mobile ions → conduct electricity; neutral NaCl solution still conducts due to ions.
Question 8
(a) Lead(II) nitrate, Pb(NO₃)₂(aq) and sodium sulfate, Na₂SO₄(aq) [1]
Or any soluble lead(II) salt (e.g., Pb(CH₃COO)₂) + any soluble sulfate (e.g., (NH₄)₂SO₄, K₂SO₄). Both must be soluble.
(b) Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s) [2]
Mark breakdown: Correct ions [1], correct product with state symbol [1]
(c) Procedure for pure, dry PbSO₄:
- Mix the two aqueous solutions in a beaker. Stir. White precipitate of PbSO₄ forms immediately. [1]
- Filter the mixture using filter paper and funnel. Collect the precipitate (residue). [1]
- Wash the residue thoroughly with distilled water to remove soluble impurities (Na⁺, NO₃⁻, etc.). [1]
- Dry the precipitate in an oven at ~100°C or between filter papers at room temperature. [1]
Total 4 marks. Key: filtration, washing with distilled water, drying. Do not evaporate/heat to dryness (insoluble salt).
(d) 2PbSO₄(s) → 2PbO(s) + 2SO₂(g) + O₂(g) [2]
Mark breakdown: Correct formulae [1], correct balancing and state symbols [1]
Question 9
(a) At equivalence point, all weak acid (CH₃COOH) is converted to its conjugate base (CH₃COO⁻). [1] The ethanoate ion undergoes hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, producing OH⁻ ions, making the solution slightly alkaline (pH > 7). [1]
(b) pKₐ = 4.76 (pH at half-equivalence point, 12.5 cm³) [1]
(c) Methyl orange changes colour at pH 3.1–4.4. [1] The equivalence point of this titration is at pH ≈ 8.7 (vertical region ~pH 7–10). Methyl orange would change colour before the equivalence point (in the buffer region), giving a falsely low titre volume. [1]
Suitable indicator: phenolphthalein (pH 8.2–10.0).
(d) At half-equivalence point, [CH₃COOH] = [CH₃COO⁻].
Using Henderson-Hasselbalch: pH = pKₐ + log([salt]/[acid]) = pKₐ + log(1) = pKₐ = 4.76 [1]
Alternatively: Kₐ = [H⁺][CH₃COO⁻]/[CH₃COOH] → [H⁺] = Kₐ = 1.74 × 10⁻⁵ → pH = -log(1.74 × 10⁻⁵) = 4.76 [2]
Mark breakdown: Recognition of half-equivalence condition [1], correct pH calculation [1], correct final answer [1]
Answer: pH = 4.76
Question 10
(a) Mᵣ of NH₄NO₃ = 14 + 4 + 14 + 48 = 80
Mass of N = 2 × 14 = 28
% N = (28 ÷ 80) × 100% = 35.0% [2]
Mark breakdown: Correct Mᵢ calculation [1], correct percentage [1]
(b)
(i) Initial moles HCl = 0.500 mol/dm³ × 0.0250 dm³ = 0.0125 mol [1]
Moles NaOH used = 0.100 mol/dm³ × 0.0185 dm³ = 0.00185 mol [1]
HCl + NaOH → NaCl + H₂O (1:1)
Moles excess HCl = moles NaOH = 0.00185 mol
Moles HCl reacted with NH₃ = 0.0125 − 0.00185 = 0.01065 mol [1]
Wait: 2 marks total. Mark breakdown: Initial moles HCl [1], moles reacted [1].
(ii) NH₃ + HCl → NH₄Cl (1:1)
Moles NH₃ = moles HCl reacted = 0.01065 mol
Each NH₄NO₃ produces 1 NH₃ → moles NH₄NO₃ = 0.01065 mol
Each NH₄NO₃ contains 2 N atoms → moles N = 2 × 0.01065 = 0.0213 mol
Mass of N = 0.0213 mol × 14 g/mol = 0.298 g [2]
Mark breakdown: Mole ratio NH₃:N [1], mass calculation [1]
(iii) Moles NH₄NO₃ = 0.01065 mol
Mass pure NH₄NO₃ = 0.01065 × 80 = 0.852 g
% purity = (0.852 ÷ 2.00) × 100% = 42.6% [2]
Mark breakdown: Mass pure salt [1], % purity calculation [1]
END OF ANSWER KEY
Total Marks: 60
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