CBSE Class 10 Chemistry Chemical Reactions and Equations MCQs Set 30

Answer: (a) (ii) Carbon dioxide is getting oxidised

Explanation:
1. Lead(II) oxide loses oxygen to carbon, so lead oxide is reduced (statements i and iv are true).
2. Carbon gains oxygen, so carbon is oxidised (iii is true).
3. Carbon dioxide is the product, not a reactant that is being oxidised, so statement (ii) is incorrect.

Teacher's Note:
1. Oxidation is gain of oxygen and reduction is loss of oxygen.
2. Always apply the terms to the reactants.

Answer: (a) oxidation

Explanation:
1. Oils and fats in food are slowly oxidised by oxygen of the air.
2. The products have an unpleasant smell and taste, which we call rancidity.

Teacher's Note:
1. Prevent rancidity with airtight containers, refrigeration or antioxidants.
2. Corrosion is the oxidation of metals.

Answer: (a) physical change

Explanation:
1. Sugar can be recovered by evaporating the water, and no new substance is formed.
2. So the change is physical.

Teacher's Note:
1. A chemical change forms new substances.
2. Dissolving is usually a physical change.

Answer: (c) Zn + 2HCl → ZnCl₂ + H₂

Explanation:
1. In Zn + 2HCl → ZnCl₂ + H₂, zinc displaces hydrogen from the acid.
2. (a) is decomposition, (b) and (d) are combination reactions.

Teacher's Note:
1. Displacement: A + BC → AC + B.
2. Zinc is more reactive than hydrogen.

Answer: (d) CaO(s) + CO₂(g) → CaCO₃(s)

Explanation:
1. In a double displacement reaction two compounds exchange their ions.
2. The first three reactions show an exchange of ions, but CaO + CO₂ → CaCO₃ is a combination reaction.

Teacher's Note:
1. Double displacement: AB + CD → AD + CB.
2. Combination: A + B → AB.

Answer: (c) double displacement reaction

Explanation:
1. The ions exchange partners: Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃.
2. A yellow precipitate of PbI₂ forms, so it is also a precipitation reaction.

Teacher's Note:
1. Pb²⁺ and K+ swap their anions.
2. Yellow PbI₂ settles out as a solid.

Answer: (a) (i), (ii) and (iv)

Explanation:
1. A species whose element is in an intermediate oxidation state can both gain and lose electrons.
2. N in HNO₂ (+3), O in H₂O₂ (–1) and S in SO₂ (+4) are intermediate.
3. S in H₂S is –2, the lowest state, so H₂S can only be a reducing agent.

Teacher's Note:
1. Highest oxidation state: only oxidising. Lowest state: only reducing.
2. Intermediate state: both.

Answer: (a) PbO is reduced and carbon acts as a reducing agent

Explanation:
1. PbO loses oxygen and Pb goes from +2 to 0, so PbO is reduced.
2. Carbon gains oxygen and goes from 0 to +2, so it is oxidised and acts as the reducing agent.

Teacher's Note:
1. This is the reduction of an ore with carbon.
2. Oxidation and reduction occur together.

Answer: (a) +1

Explanation:
1. Water is neutral and NO is present as NO+, so the complex charge +2 = x + 0 + 1.
2. Therefore x = +1.

Teacher's Note:
1. The brown ring test is used for nitrate.
2. Here Fe is in an unusual +1 state.

Answer: (c) H₂S + 8HNO₃ → H₂SO₄ + 8NO₂ + 4H₂O

Explanation:
1. Sulphur goes from –2 to +6 (8 electrons lost) and each N goes from +5 to +4 (1 electron gained).
2. So 8 HNO₃ are reduced for each H₂S.
3. Atoms then balance: H 10, O 24, N 8 on both sides.

Teacher's Note:
1. The number of electrons lost by S fixes the coefficient of NO₂.
2. Use H₂O to balance hydrogen and oxygen last.

Answer: (b) ferric oxide

Explanation:
1. On heating, FeSO₄·7H₂O first loses water, then the anhydrous salt decomposes: 2FeSO₄ → Fe₂O₃ + SO₂ + SO₃.
2. The solid residue is red-brown ferric oxide.

Teacher's Note:
1. The green crystals turn reddish brown.
2. The gases SO₂ and SO₃ have a smell of burning sulphur.

Answer: (d) Total change in oxidation number of iodine is 10 and of oxygen is 2

Explanation:
1. I goes from 0 in I₂ to +5 in HIO₃, a change of 5 per atom and 10 for two atoms.
2. Oxygen goes from –1 in H₂O₂ to –2 in H₂O, a change of 1 per atom and 2 for the two oxygen atoms of H₂O₂.

Teacher's Note:
1. Total change = change per atom × number of atoms.
2. Electrons lost and gained must match: 5 H₂O₂ per I₂.

Answer: (b) 1.33

Explanation:
1. Moles of Fe(OH)₃ from each reactant: Fe₂S₃ gives 1 × 4/2 = 2, H₂O gives 2 × 4/6 = 1.33, O₂ gives 3 × 4/3 = 4.
2. The limiting reagent is water, so 1.33 mol of Fe(OH)₃ is formed.

Teacher's Note:
1. The reactant giving the least product is limiting.
2. Compare moles using the coefficients.

Answer: (c) formation of a precipitate and evolution of a gas

Explanation:
1. A downward arrow shows that the product is formed as an insoluble solid (precipitate).
2. An upward arrow shows that the product escapes as a gas.

Teacher's Note:
1. Examples: BaSO₄↓ and H₂↑.
2. State symbols (s), (l), (g), (aq) are also used.

Answer: (b) they are very unreactive and do not react with oxygen or moisture

Explanation:
1. Gold and silver are placed very low in the reactivity series.
2. They do not react readily with air and water, so they do not corrode (silver only tarnishes slowly with sulphur).

Teacher's Note:
1. Corrosion is the slow eating away of a metal by air and moisture.
2. Iron, copper and silver are the metals that corrode.

Answer: (a) a displacement reaction and a combination reaction

Explanation:
1. In (a) chlorine displaces iodine from potassium iodide.
2. In (b) two elements combine to form one compound.

Teacher's Note:
1. Displacement: one element replaces another in a compound.
2. Combination: two or more substances form one product.

Answer: (a) No change takes place, as silver is less reactive than copper

Explanation:
1. Silver is below copper in the reactivity series, so it cannot displace copper from its salt.
2. So there is no reaction.

Teacher's Note:
1. The solution stays blue.
2. A more reactive metal is needed for displacement.

Answer: (d) yellow lead(II) iodide: Pb(NO₃)₂ + 2KI → PbI₂↓ + 2KNO₃

Explanation:
1. The ions exchange partners and lead iodide is insoluble.
2. PbI₂ is a bright yellow precipitate.

Teacher's Note:
1. This reaction is a double displacement and a precipitation reaction.
2. The yellow solid can be dissolved on heating and reappears on cooling as golden spangles.

Answer: (a) 2Fe(s) + 3H₂O(g) → Fe₂O₃(s) + 3H₂(g)

Explanation:
1. Balance Fe: 2 Fe on the left for Fe₂O₃ on the right.
2. Oxygen then fixes 3 H₂O, giving 3 H₂.
3. 2Fe(s) + 3H₂O(g) → Fe₂O₃(s) + 3H₂(g) has equal atoms on both sides.

Teacher's Note:
1. Steam is written as (g).
2. Check each element after balancing.

Answer: (b) prevent contact with air and moisture, and so prevent rusting

Explanation:
1. Paint covers the surface and keeps oxygen and moisture away from the iron.
2. Without both, rusting cannot take place.

Teacher's Note:
1. Other methods: galvanising, oiling, greasing, alloying (stainless steel).
2. Rusting weakens iron structures.

Answer: (b) metals (for example rusting of iron) and fats and oils in food

Explanation:
1. Corrosion is the slow eating away of a metal by air, moisture or chemicals, for example rusting of iron.
2. Rancidity is the oxidation of fats and oils in food, giving a bad smell and taste.

Teacher's Note:
1. Both are slow oxidation reactions.
2. Prevent corrosion by painting; prevent rancidity with nitrogen flushing.

Answer: (a) 10

Explanation:
1. Balance the atoms of each element in turn.
2. The balanced equation is 2Al + 3CuCl₂ → 2AlCl₃ + 3Cu.
3. The coefficients add up to 2+3+2+3 = 10.

Teacher's Note:
1. Never change a subscript while balancing, only the coefficients.
2. Check every element on both sides at the end.

Answer: (a) 8

Explanation:
1. Balance the atoms of each element in turn.
2. The balanced equation is SO₂ + 2H₂S → 2H₂O + 3S.
3. The coefficients add up to 1+2+2+3 = 8.

Teacher's Note:
1. Never change a subscript while balancing, only the coefficients.
2. Check every element on both sides at the end.