Chapter 4 · Solved paper40 marks

Model Test – Solved ഉത്തരങ്ങൾ

Every question of the model paper, in the same order, with the full answer, the marking scheme, an animated step-by-step solution for the longer questions, and a link to the page where the idea is taught. Try each one again before you open the answer, and mark your own paper honestly.

📐 How the solutions are drawn

In equations a coefficient that has just been changed is red, as in the textbook. The atom tally under an equation turns green when an atom is equal on both sides. Mass bars show the masses in u side by side – when the reactant bar and the product bar end at the same place, mass is conserved. Oxidation numbers are written above the symbols; a blue arrow above the equation shows an increase (oxidation) and a red arrow below it a decrease (reduction).

Rules used everywhere: total mass of reactants = total mass of products; balance with coefficients only; loss of electrons = oxidation = oxidation number up; gain = reduction = down; the oxidising agent is reduced and the reducing agent oxidised; free elements are 0, the sum in a compound is 0 (H +1, O −2, Na and K +1, Mg and Ca +2, Al +3).

SECTION AAnswer all · 8 × 1 = 8 marks

A 2.4 g piece of magnesium ribbon burns completely in air and forms 4.0 g of magnesium oxide. The mass of oxygen that combined with the magnesium is

(a) 6.4 g(b) 1.6 g(c) 2.4 g(d) 4.0 g

(1 mark)

💡 Total mass of reactants = total mass of products.

✅ Answer: (b) 1.6 g

Law of conservation of mass (മാസ് സംരക്ഷണനിയമം): mass of magnesium + mass of oxygen = mass of magnesium oxide → 2.4 g + ? = 4.0 g → oxygen = 4.0 − 2.4 = 1.6 g. (Check with 2Mg + O₂ → 2MgO: 48 u of Mg combine with 32 u of O₂; 48 : 32 = 2.4 : 1.6.)

Marking: 1 mark for (b).

📖 Review this on Page 3 · Conservation of mass

How many atoms in all are there in 3Ca(OH)₂?

(1 mark)

💡 First count the atoms in one unit, then multiply by the coefficient.

✅ Answer: 15 atoms

One Ca(OH)₂ has 1 Ca + 2 O + 2 H = 5 atoms (the 2 outside the bracket multiplies both O and H). The coefficient 3 means three units: 3 × 5 = 15 atoms (Ca 3, O 6, H 6).

Marking: 1 mark.

📖 Review this on Page 1 · Reading equations · Page 4 · Balancing equations

The oxidation number of phosphorus in a P₄ molecule is

(a) +4(b) −3(c) 0(d) +5

(1 mark)

💡 No other element is present – who could take the electrons?

✅ Answer: (c) 0

P₄ is a molecule of an element – phosphorus is in the elemental state (മൂലകാവസ്ഥ). The oxidation number of an element in the elemental state is 0, however many atoms the molecule has (like H₂, O₂, Cl₂).

Marking: 1 mark for (c).

📖 Review this on Page 7 · Oxidation number

Find the relation and fill in: Na → Na⁺ + e⁻ : oxidation :: Zn²⁺ + 2e⁻ → Zn :

(1 mark)

💡 Are electrons given out or taken in?

✅ Answer: reduction (നിരോക്സീകരണം)

Sodium loses an electron – oxidation. The zinc ion gains two electrons and becomes a zinc atom – gain of electrons is reduction (positive ions that gain electrons are reduced, like Ag⁺ + e⁻ → Ag).

Marking: 1 mark.

📖 Review this on Page 5 · Oxidation & reduction

In the reaction CuO + H₂ → Cu + H₂O, the reducing agent is

(a) CuO(b) H₂(c) Cu(d) H₂O

(1 mark)

💡 The reducing agent is itself oxidised.

✅ Answer: (b) H₂

Cu+2O−2 + H0₂ → Cu0 + H+1₂O−2. Hydrogen: 0 → +1 – it is oxidised, so H₂ is the reducing agent (നിരോക്സീകാരി). Copper: +2 → 0 – CuO is reduced; it is the oxidising agent.

Marking: 1 mark for (b).

📖 Review this on Page 6 · Oxidising & reducing agents · Page 8 · Redox reactions

The oxidation number of nitrogen in ammonia, NH₃, is .

(1 mark)

💡 H is +1; the total is zero.

✅ Answer: −3

Nitrogen is more electronegative than hydrogen, so H is +1. Sum of oxidation numbers = 0: x + 3(+1) = 0 → x + 3 = 0 → x = −3. So N is −3.

Marking: 1 mark (the sign is needed).

📖 Review this on Page 7 · Oxidation number

Which of these is not a redox reaction?

(a) 2Mg + O₂ → 2MgO(b) 2Na + 2H₂O → 2NaOH + H₂(c) KOH + HNO₃ → KNO₃ + H₂O(d) 2H₂ + O₂ → 2H₂O

(1 mark)

💡 Look for an element whose oxidation number changes.

✅ Answer: (c) KOH + HNO₃ → KNO₃ + H₂O

K+1O−2H+1 + H+1N+5O−2₃ → K+1N+5O−2₃ + H+1₂O−2: K +1, O −2, H +1, N +5 – no oxidation number changes, so it is not a redox reaction (it is a neutralisation). In the other three a metal or hydrogen goes from 0 to a positive value (oxidation) while another element is reduced.

Marking: 1 mark for (c).

📖 Review this on Page 8 · Redox reactions

Write the missing coefficient so that the equation is balanced: 2KClO₃ → 2KCl + O₂

(1 mark)

💡 Count the oxygen atoms on the left.

✅ Answer: 3

Left: 2 × 3 = 6 oxygen atoms. Right: each O₂ has 2, so 6 ÷ 2 = 3. 2KClO₃ → 2KCl + 3O₂ – K 2 = 2, Cl 2 = 2, O 6 = 6.

Marking: 1 mark.

📖 Review this on Page 4 · Balancing equations

SECTION BAnswer all · 5 × 2 = 10 marks

Sea shells, which are mostly calcium carbonate, are heated in kilns to make lime: CaCO₃ → CaO + CO₂. Using the atomic masses Ca = 40, C = 12, O = 16, show that this equation obeys the law of conservation of mass.

(2 marks)

💡 Add the masses of all the atoms in each formula.

🎬 Step by step

✅ Answer: reactants 100 u = products 100 u

Reactant: CaCO₃ = 40 + 12 + 3 × 16 = 100 u.

Products: CaO = 40 + 16 = 56 u; CO₂ = 12 + 2 × 16 = 44 u; total 56 + 44 = 100 u.

Total mass of reactants = total mass of products (ആകെ മാസ്) = 100 u, so the law of conservation of mass is obeyed. The atoms are only rearranged.

Marking: reactant mass 1 mark; product masses and conclusion 1 mark.

📖 Review this on Page 3 · Conservation of mass

10.

Balance the following equations. (a) Al + HCl → AlCl₃ + H₂ (b) C₃H₈ + O₂ → CO₂ + H₂O (propane, one of the gases in LPG)

(2 marks)

💡 Balance the atom that appears in only one substance on each side first; leave O₂ (or H₂) for last.

🎬 Step by step

✅ Answer: (a) 2Al + 6HCl → 2AlCl₃ + 3H₂ (b) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

(a) Chlorine: 3 in AlCl₃ → 3HCl. Then H is 3 on the left but 2 in H₂; make H even: 6HCl → 3H₂, and then 2AlCl₃ and 2Al: 2Al + 6HCl → 2AlCl₃ + 3H₂

Atom	Reactants	Products
Al	2	2
H	6	6
Cl	6	6

(b) Carbon: 3 → 3CO₂. Hydrogen: 8 → 4H₂O. Oxygen on the right: 3 × 2 + 4 × 1 = 10 → 5O₂: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Atom	Reactants	Products
C	3	3
H	8	8
O	10	10

Only the coefficients are changed – never the subscripts.

Marking: 1 mark for each balanced equation.

📖 Review this on Page 4 · Balancing equations

11.

Potassium reacts with bromine: 2K + Br₂ → 2KBr. (a) Write the equations showing the change that happens to a potassium atom and to a bromine atom. (b) Which of these changes is oxidation and which is reduction? Give the reason.

(2 marks)

💡 Metals lose electrons; non-metals gain them.

🎬 Step by step

✅ Answer: (a) K → K⁺ + e⁻; Br + e⁻ → Br⁻ (b) potassium: oxidation; bromine: reduction

(a) Potassium (2, 8, 8, 1), a metal, gives away its outermost electron: K → K⁺ + e⁻. Bromine, a non-metal like chlorine, takes one electron: Br + e⁻ → Br⁻. The ions form KBr.

(b) Potassium loses an electron → oxidation (ഓക്സീകരണം). Bromine gains an electron → reduction (നിരോക്സീകരണം). The 2 electrons lost by two potassium atoms are gained by the two bromine atoms of one Br₂ molecule.

Marking: (a) ½ + ½; (b) ½ + ½ (with the reason).

📖 Review this on Page 5 · Oxidation & reduction

12.

Find the oxidation number of (a) chlorine in sodium hypochlorite, NaClO (used in bleaching liquids) (b) phosphorus in phosphoric acid, H₃PO₄. Show the working.

(2 marks)

💡 Write x for the unknown and make the total zero.

🎬 Step by step

✅ Answer: (a) Cl +1 (b) P +5

(a) Na = +1, O = −2, Cl = x: (+1) + x + (−2) = 0 → 1 + x − 2 = 0 → x = +1. Na+1Cl+1O−2

(b) H = +1, O = −2, P = x: 3(+1) + x + 4(−2) = 0 → 3 + x − 8 = 0 → x = +5. H+1₃P+5O−2₄

Chlorine is −1 only when it is the more electronegative partner (NaCl, HCl); here oxygen is −2 and chlorine takes the value that makes the sum zero.

Marking: 1 mark each (working ½, answer ½).

📖 Review this on Page 7 · Oxidation number

13.

In the thermite reaction, used to weld railway tracks, aluminium reacts with iron oxide: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe. Using oxidation numbers, find (a) the element oxidised and the element reduced (b) the oxidising agent and the reducing agent.

(2 marks)

💡 Write the oxidation number above every atom first.

🎬 Step by step

✅ Answer: (a) Al oxidised (0 → +3), Fe reduced (+3 → 0) (b) oxidising agent Fe₂O₃, reducing agent Al

2Al0 + Fe+3₂O−2₃ → Al+3₂O−2₃ + 2Fe0

(a) Aluminium: 0 → +3 – the oxidation number increases → aluminium is oxidised. Iron: +3 → 0 – it decreases → iron is reduced. Oxygen stays −2.

(b) Fe₂O₃ makes aluminium’s oxidation number rise and is itself reduced → oxidising agent (ഓക്സീകാരി). Aluminium is itself oxidised → reducing agent (നിരോക്സീകാരി).

Marking: (a) ½ + ½; (b) ½ + ½.

📖 Review this on Page 8 · Redox reactions · Page 6 · Oxidising & reducing agents

SECTION CAnswer all · 4 × 3 = 12 marks

14.

Iron is extracted from its oxide in a blast furnace: Fe₂O₃ + CO → Fe + CO₂. (a) Balance the equation.
(b) Using the atomic masses Fe = 56, C = 12, O = 16, show that the total mass is the same on both sides of the balanced equation.
(c) Which substance is reduced in this reaction? Give the reason using oxidation numbers.

(3 marks)

💡 Count the oxygen atoms that CO can take from Fe₂O₃.

🎬 Step by step

✅ Answer: (a) Fe₂O₃ + 3CO → 2Fe + 3CO₂ (b) 244 u = 244 u (c) Fe₂O₃ (Fe +3 → 0)

(a) Fe: 2 → 2Fe. O: Fe₂O₃ gives 3 oxygen atoms, and each CO takes one to become CO₂ → 3CO and 3CO₂: Fe₂O₃ + 3CO → 2Fe + 3CO₂

Atom	Reactants	Products
Fe	2	2
O	6	6
C	3	3

(b) Reactants: Fe₂O₃ = 2 × 56 + 3 × 16 = 160 u; 3CO = 3 × (12 + 16) = 84 u; total 244 u.
Products: 2Fe = 2 × 56 = 112 u; 3CO₂ = 3 × (12 + 2 × 16) = 132 u; total 244 u. Equal – mass is conserved.

Fe+3₂O−2₃ + 3C+2O−2 → 2Fe0 + 3C+4O−2₂

Marking: 1 mark for each part.

📖 Review this on Page 4 · Balancing equations · Page 3 · Conservation of mass · Page 8 · Redox reactions

15.

Find the oxidation number of (a) manganese in potassium manganate, K₂MnO₄
(b) nitrogen in dinitrogen trioxide, N₂O₃
(c) sulphur in aluminium sulphate, Al₂(SO₄)₃.

(3 marks)

💡 In (c) the bracket multiplies: 3 sulphur atoms and 12 oxygen atoms.

🎬 Step by step

✅ Answer: (a) +6 (b) +3 (c) +6

(a) K = +1, O = −2, Mn = x: 2(+1) + x + 4(−2) = 0 → 2 + x − 8 = 0 → x = +6 K+1₂Mn+6O−2₄

(b) O = −2, N = x: 2x + 3(−2) = 0 → 2x − 6 = 0 → 2x = 6 → x = +3 N+3₂O−2₃

(c) Al = +3, O = −2; there are 3 S and 3 × 4 = 12 O atoms: 2(+3) + 3x + 12(−2) = 0 → 6 + 3x − 24 = 0 → 3x = 18 → x = +6 Al+3₂(S+6O−2₄)₃

Always multiply each oxidation number by the number of atoms of that element.

Marking: 1 mark each (working ½, answer with sign ½).

📖 Review this on Page 7 · Oxidation number

16.

Write the oxidation number of every atom above its symbol and find out whether each reaction is a redox reaction. (a) CuO + 2HCl → CuCl₂ + H₂O
(b) 2FeCl₂ + Cl₂ → 2FeCl₃
(c) For the redox reaction, name the oxidising agent and the reducing agent.

(3 marks)

💡 Elements in the free state are 0.

🎬 Step by step

✅ Answer: (a) not redox (b) redox (c) oxidising agent Cl₂, reducing agent FeCl₂

(a) Cu+2O−2 + 2H+1Cl−1 → Cu+2Cl−1₂ + H+1₂O−2

Cu +2 → +2, O −2 → −2, H +1 → +1, Cl −1 → −1. No oxidation number changes → not a redox reaction.

(b) 2Fe+2Cl−1₂ + Cl0₂ → 2Fe+3Cl−1₃

Iron: +2 → +3 (increases → oxidised). Chlorine of Cl₂: 0 → −1 (decreases → reduced). Oxidation and reduction happen together → a redox reaction.

Marking: (a) 1 mark; (b) 1 mark; (c) ½ + ½.

📖 Review this on Page 8 · Redox reactions

17.

Anu puts about 20 mL of dilute hydrochloric acid in a conical flask. She places a few pieces of magnesium ribbon in a balloon, ties the balloon over the mouth of the flask without letting the magnesium fall in, and keeps the flask on an electronic balance. The reading is 86.40 g. She then lifts the balloon so that the magnesium falls into the acid. (a) Write two observations.
(b) Write the balanced equation of the reaction. What will the balance show after the reaction? Why?
(c) When she repeats the experiment in an open flask without the balloon, the reading becomes less. Does this go against the law of conservation of mass? Explain.

(3 marks)

💡 Is anything able to get out of the flask?

🎬 Step by step

✅ Answer: (a) bubbles, balloon swells (b) Mg + 2HCl → MgCl₂ + H₂; still 86.40 g (c) no – hydrogen escapes

(a) Bubbles of gas form (effervescence) and the balloon swells; the magnesium pieces slowly disappear and the flask becomes warm.

(b) Mg + 2HCl → MgCl₂ + H₂↑. The reading stays 86.40 g: the flask is closed by the balloon, so the hydrogen formed stays inside – nothing enters or leaves. The total mass of the products equals the total mass of the reactants.

(c) No. In the open flask the hydrogen gas escapes into the air, so the balance does not weigh it. If the escaping gas were collected and weighed, the total mass would still be the same – just as burning wood seems to lose mass only because CO₂ and water vapour escape.

Marking: (a) ½ + ½; (b) equation ½, reading with the reason ½; (c) 1 mark.

📖 Review this on Page 3 · Conservation of mass

SECTION DAnswer all · 2 × 5 = 10 marks

18.

Burning magnesium keeps burning even in carbon dioxide, forming white magnesium oxide and black specks of carbon: Mg + CO₂ → MgO + C. (a) Balance the equation.
(b) Using the atomic masses Mg = 24, C = 12, O = 16, show that mass is conserved.
(c) Write the oxidation numbers of Mg, C and O before and after the reaction.
(d) Name the element oxidised, the element reduced, the oxidising agent and the reducing agent.
(e) Why should a carbon dioxide fire extinguisher not be used on burning magnesium?

(5 marks)

💡 Do (a) first – parts (b) to (d) all use the balanced equation.

🎬 Step by step

✅ Answer: (a) 2Mg + CO₂ → 2MgO + C (b) 92 u = 92 u (c) Mg 0 → +2, C +4 → 0, O −2 → −2 (d) Mg oxidised, C reduced; oxidising agent CO₂, reducing agent Mg (e) Mg takes the oxygen of CO₂

(a) O: 2 in CO₂ → 2MgO, then 2Mg: 2Mg + CO₂ → 2MgO + C

Atom	Reactants	Products
Mg	2	2
C	1	1
O	2	2

(b) Reactants: 2Mg = 2 × 24 = 48 u; CO₂ = 12 + 2 × 16 = 44 u; total 92 u. Products: 2MgO = 2 × (24 + 16) = 80 u; C = 12 u; total 92 u. Equal.

Mg: 0 → +2; C: +4 → 0 (in CO₂: x + 2(−2) = 0 → x = +4); O: −2 → −2 (no change).

(d) Magnesium (0 → +2) is oxidised; carbon (+4 → 0) is reduced. CO₂ is the oxidising agent; magnesium is the reducing agent.

(e) Magnesium is such a strong reducing agent that it pulls the oxygen out of CO₂ and goes on burning – the extinguisher would feed the fire. A magnesium fire is smothered with dry sand.

Marking: 1 mark for each part ((c) ½ for Mg and C, ½ for O; (d) ½ for oxidised/reduced, ½ for the two agents).

📖 Review this on Page 4 · Balancing equations · Page 3 · Conservation of mass · Page 7 · Oxidation number · Page 8 · Redox reactions

19.

Read the passage and answer the questions.

At Neendakara fishing harbour in Kollam, Rafeeq’s family owns a fishing boat. Its engine frame and anchor are made of iron. In the moist, salty sea air the iron parts are soon covered with a reddish-brown layer of rust, and the rusted anchor becomes heavier than when it was new. Rust forms when iron reacts with the oxygen and water of the air; in this question take rust as iron(III) hydroxide, Fe(OH)₃. Rafeeq’s father paints the iron parts before every monsoon. The roof of their house is made of galvanised iron (GI) sheets – iron sheets coated with zinc. The zinc slowly forms a thin, white oxide layer (ഓക്സൈഡ് ആവരണം), and the sheets do not rust for many years.

(a) Balance the equation for rusting: Fe + O₂ + H₂O → Fe(OH)₃.
(b) Find the oxidation number of iron before and after rusting. Write the equation for the change of an iron atom. Is iron oxidised or reduced?
(c) Which substance is the oxidising agent? Write the equation for the change of an oxygen atom.
(d) Does the increase in mass of the rusted anchor go against the law of conservation of mass? Explain.
(e) On the GI sheet, zinc reacts with oxygen: 2Zn + O₂ → 2ZnO. Write the change in the oxidation number of zinc and explain how the zinc coating protects the iron.

(5 marks)

💡 Underline the substances in the passage; write the oxidation numbers above them.

🎬 Step by step

✅ Answer: (a) 4Fe + 3O₂ + 6H₂O → 4Fe(OH)₃ (b) 0 → +3; Fe → Fe³⁺ + 3e⁻; oxidised (c) O₂; O + 2e⁻ → O²⁻ (d) no (e) 0 → +2; zinc is oxidised instead of iron

(a) Start with H (2 in H₂O, 3 in Fe(OH)₃): 2Fe(OH)₃ has 6 H → 3H₂O. Its 6 O atoms = 3 from water + 3 → 1½O₂. Double everything to remove the half: 4Fe + 3O₂ + 6H₂O → 4Fe(OH)₃

Atom	Reactants	Products
Fe	4	4
O	12	12
H	12	12

4Fe0 + 3O0₂ + 6H+1₂O−2 → 4Fe+3(O−2H+1)₃

(b) Iron is 0 in the free metal. In Fe(OH)₃: x + 3(−2) + 3(+1) = 0 → x − 6 + 3 = 0 → x = +3. So 0 → +3: Fe → Fe³⁺ + 3e⁻ – loss of electrons, an increase in oxidation number → iron is oxidised.

(d) No. The anchor gains the mass of the oxygen and water from the air that combined with the iron. Mass of iron + oxygen + water used = mass of rust formed – the total mass does not change. In u: 4Fe = 4 × 56 = 224 u; 3O₂ = 3 × (2 × 16) = 96 u; 6H₂O = 6 × (2 × 1 + 16) = 108 u; total 428 u = 4Fe(OH)₃ = 4 × (56 + 3 × 16 + 3 × 1) = 428 u.

(e) 2Zn0 + O0₂ → 2Zn+2O−2: zinc 0 → +2 – zinc is oxidised. Zinc is oxidised more readily than iron, so the air and moisture react with the zinc instead of the iron, and the thin zinc oxide layer it forms covers the sheet and keeps air and water away from the iron (paint does the same job on the boat).

Marking: 1 mark for each part (½ + ½ where there are two answers).

📖 Review this on Page 4 · Balancing equations · Page 8 · Redox reactions · Page 5 · Oxidation & reduction · Page 3 · Conservation of mass · Page 11 · Applications

Marks summary

Section	Questions	Marks each	Total	Topics
A	1–8	1	8	mass of oxygen in burning Mg, atoms in 3Ca(OH)₂, oxidation number of P₄, oxidation : reduction, reducing agent, N in NH₃, not redox, balancing KClO₃
B	9–13	2	10	conservation of mass for CaCO₃ in u, balancing Al + HCl and propane, half-equations for KBr, oxidation numbers in NaClO and H₃PO₄, thermite reaction
C	14–17	3	12	blast furnace (balance, mass, reduction), oxidation numbers in K₂MnO₄, N₂O₃, Al₂(SO₄)₃, redox or not, the balloon-and-balance experiment
D	18–19	5	10	magnesium burning in CO₂, Kerala case study on the rusting fishing boat and GI sheets
Total			40

⏱️ Time management (15 min cool-off + 90 minutes)

Cool-off (15 min) – read the whole paper. Mark the equations to balance (Q8, Q10, Q14, Q18, Q19) and the questions that need oxidation numbers (Q3, Q5, Q6, Q7, Q12, Q13, Q14, Q15, Q16, Q18, Q19).
Section A – about 10 minutes. One word, number or letter each.
Section B – about 20 minutes. Show the working: the mass of each formula, the atom count, the “sum = 0” line.
Section C – about 25 minutes. Write the oxidation number above every atom before you decide what is oxidised or reduced.
Section D – about 25 minutes. Balance first – every later part uses the balanced equation. In Q19 underline the substances in the passage.
Checking – keep 10 minutes: every equation balanced atom by atom? Signs on all oxidation numbers? Oxidising agent = the substance reduced?

🎉 Finished the chapter?

Count your marks: 32 or more – excellent; 24–31 – good, revise the pages linked under the questions you missed; below 24 – go back through pages 3 to 8 with their labs and do the scored quiz again. Then return to the chapter home and tick off what you have learned.