Chemical Reaction and Equations — NCERT Solutions

(i) The substance is calcium oxide or quick lime; formula: CaO.

(ii) Its reaction with water is:

$$\text{CaO(s)} + \text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(aq)} + \text{heat}$$

Given: Magnesium ribbon burns in air (oxygen) to form magnesium oxide.

Reaction:
$$2\mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{MgO}(\mathrm{s})$$

Concept: A substance is said to be oxidised if it gains oxygen during a reaction.

Conclusion: Magnesium gains oxygen to form magnesium oxide. Therefore, magnesium is being oxidised in this reaction.

Given: An iron nail is dipped in copper sulphate (CuSO₄) solution.

Concept: This is a displacement reaction. A more reactive metal displaces a less reactive metal from its salt solution. Iron is more reactive than copper.

Reaction:
$$\mathrm{Fe}(\mathrm{s}) + \mathrm{CuSO}_4(\mathrm{aq}) \rightarrow \mathrm{FeSO}_4(\mathrm{aq}) + \mathrm{Cu}(\mathrm{s})$$

Explanation: Iron displaces copper from copper sulphate solution. The blue colour of copper sulphate solution fades (or turns light green) because blue CuSO₄ is consumed and green FeSO₄ is formed. A reddish-brown deposit of copper metal appears on the iron nail.

Answer: The colour of copper sulphate solution changes because iron displaces copper from the solution, forming iron sulphate (FeSO₄) which is light green, and copper metal is deposited on the nail.

Concept: In a double displacement reaction, two compounds react by exchanging their ions (or groups of atoms) to form two new compounds.

Example: Reaction between sodium sulphate and barium chloride:
$$\mathrm{Na}_2\mathrm{SO}_4(\mathrm{aq}) + \mathrm{BaCl}_2(\mathrm{aq}) \rightarrow \mathrm{BaSO}_4(\mathrm{s})\downarrow + 2\mathrm{NaCl}(\mathrm{aq})$$

Explanation: Here, $\mathrm{Na}^+$ and $\mathrm{Ba}^{2+}$ ions exchange their respective anions. Barium sulphate ($\mathrm{BaSO}_4$) is formed as a white insoluble precipitate. This is also a precipitation reaction.

Concept:

• Oxidation = gain of oxygen OR loss of hydrogen.
• Reduction = loss of oxygen OR gain of hydrogen.

(i) $4\mathrm{Na}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{Na}_2\mathrm{O}(\mathrm{s})$

• Sodium (Na) gains oxygen to form sodium oxide (Na₂O). Therefore, sodium (Na) is oxidised.
• Oxygen ($\mathrm{O}_2$) itself is the oxidising agent; it is being used up to oxidise Na. (In this reaction, only oxidation occurs — there is no separate substance being reduced in the classical sense, but $\mathrm{O}_2$ acts as the oxidising agent.)

Answer: Sodium (Na) is oxidised (gains oxygen).

(ii) $\mathrm{CuO(s)} + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{Cu(s)} + \mathrm{H}_2\mathrm{O(l)}$

• CuO loses oxygen to form Cu. Therefore, CuO (copper oxide) is reduced.
• H₂ gains oxygen to form H₂O. Therefore, hydrogen (H₂) is oxidised.

Answer:

• Substance oxidised → $\mathrm{H}_2$ (hydrogen gas gains oxygen).
• Substance reduced → $\mathrm{CuO}$ (copper oxide loses oxygen).

Correct Option: (i) (a) and (b)

Analysis of the reaction:
$$2\mathrm{PbO}(\mathrm{s}) + \mathrm{C}(\mathrm{s}) \rightarrow 2\mathrm{Pb}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})$$

• PbO loses oxygen → PbO is reduced (statement (d) is correct).
• Pb is the product of reduction of PbO, so lead itself is not getting reduced — statement (a) is incorrect.
• Carbon (C) gains oxygen to form CO₂ → Carbon is oxidised (statement (c) is correct).
• CO₂ is the product of oxidation; it is not getting oxidised further — statement (b) is incorrect.

Incorrect statements: (a) and (b) → Option (i)

Correct Option: (d) displacement reaction.

Justification: Aluminium (Al) is more reactive than iron (Fe). It displaces iron from iron(III) oxide (Fe₂O₃), taking its place. A single more reactive element displaces a less reactive element from its compound — this is the definition of a displacement reaction. This reaction is also known as the thermite reaction.

Correct Answer: (a) Hydrogen gas and iron chloride are produced.

Reaction:
$$\mathrm{Fe}(\mathrm{s}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{FeCl}_2(\mathrm{aq}) + \mathrm{H}_2(\mathrm{g})\uparrow$$

Justification: Iron is more reactive than hydrogen. It displaces hydrogen from dilute hydrochloric acid, forming iron(II) chloride (FeCl₂) and hydrogen gas. This is a displacement reaction.

Balanced Chemical Equation:

A balanced chemical equation is one in which the number of atoms of each element is equal on both the reactant side and the product side of the equation.

Example:
$$2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$$
Here, H: 4 on each side; O: 2 on each side — the equation is balanced.

Why should chemical equations be balanced?

Chemical equations must be balanced to satisfy the Law of Conservation of Mass, which states that matter can neither be created nor destroyed in a chemical reaction. This means the total mass of reactants must equal the total mass of products. Since atoms are neither created nor destroyed during a reaction, the number of atoms of each element must be the same on both sides of the equation. An unbalanced equation violates this fundamental law.

(a) Hydrogen gas combines with nitrogen to form ammonia.

Unbalanced: $\mathrm{H}_2 + \mathrm{N}_2 \rightarrow \mathrm{NH}_3$

Balancing:

• N: 2 on left, 1 on right → put 2 before NH₃: $\mathrm{H}_2 + \mathrm{N}_2 \rightarrow 2\mathrm{NH}_3$
• H: 2 on left, 6 on right → put 3 before H₂: $3\mathrm{H}_2 + \mathrm{N}_2 \rightarrow 2\mathrm{NH}_3$

Balanced equation:
$$3\mathrm{H}_2(\mathrm{g}) + \mathrm{N}_2(\mathrm{g}) \rightarrow 2\mathrm{NH}_3(\mathrm{g})$$

(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.

Unbalanced: $\mathrm{H}_2\mathrm{S} + \mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} + \mathrm{SO}_2$

Balancing:

• S: 1 on each side ✓
• H: 2 on each side ✓
• O: 2 on left, 3 on right → Balance O by adjusting coefficients.
• Try: $2\mathrm{H}_2\mathrm{S} + 3\mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + 2\mathrm{SO}_2$
• S: 2 = 2 ✓; H: 4 = 4 ✓; O: 6 = 6 ✓

Balanced equation:
$$2\mathrm{H}_2\mathrm{S}(\mathrm{g}) + 3\mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{H}_2\mathrm{O}(\mathrm{l}) + 2\mathrm{SO}_2(\mathrm{g})$$

(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.

Unbalanced: $\mathrm{BaCl}_2 + \mathrm{Al}_2(\mathrm{SO}_4)_3 \rightarrow \mathrm{AlCl}_3 + \mathrm{BaSO}_4$

Balancing:

• Al: 2 on left → put 2 before AlCl₃
• SO₄: 3 on left → put 3 before BaSO₄
• Ba: 3 on right → put 3 before BaCl₂
• Cl: 6 on left (3×2) → 6 on right (2×3) ✓

Balanced equation:
$$3\mathrm{BaCl}_2(\mathrm{aq}) + \mathrm{Al}_2(\mathrm{SO}_4)_3(\mathrm{aq}) \rightarrow 2\mathrm{AlCl}_3(\mathrm{aq}) + 3\mathrm{BaSO}_4(\mathrm{s})\downarrow$$

Check: Ba=3, Cl=6, Al=2, S=3, O=12 — equal on both sides ✓

(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

Unbalanced: $\mathrm{K} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{KOH} + \mathrm{H}_2$

Balancing:

• K: 1 on each side ✓
• O: 1 on each side ✓
• H: 2 on left, 3 on right → multiply through:
• $2\mathrm{K} + 2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{KOH} + \mathrm{H}_2$
• K: 2=2 ✓; O: 2=2 ✓; H: 4 left = 2+2=4 right ✓

Balanced equation:
$$2\mathrm{K}(\mathrm{s}) + 2\mathrm{H}_2\mathrm{O}(\mathrm{l}) \rightarrow 2\mathrm{KOH}(\mathrm{aq}) + \mathrm{H}_2(\mathrm{g})\uparrow$$

(a) $\mathrm{HNO}_3 + \mathrm{Ca(OH)}_2 \rightarrow \mathrm{Ca(NO}_3)_2 + \mathrm{H}_2\mathrm{O}$

• $\mathrm{Ca(NO}_3)_2$ has 2 NO₃ groups → need 2 HNO₃ on left.
• H on left: 2(from HNO₃) + 2(from Ca(OH)₂) = 4 → need 2 H₂O on right.

Balanced:
$$2\mathrm{HNO}_3(\mathrm{aq}) + \mathrm{Ca(OH)}_2(\mathrm{aq}) \rightarrow \mathrm{Ca(NO}_3)_2(\mathrm{aq}) + 2\mathrm{H}_2\mathrm{O}(\mathrm{l})$$
Check: H=4, N=2, O=8, Ca=1 — equal on both sides ✓

(b) $\mathrm{NaOH} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + \mathrm{H}_2\mathrm{O}$

• Na₂SO₄ needs 2 Na → 2 NaOH on left.
• H on left: 2(NaOH) + 2(H₂SO₄) = 4 → 2 H₂O on right.

Balanced:
$$2\mathrm{NaOH}(\mathrm{aq}) + \mathrm{H}_2\mathrm{SO}_4(\mathrm{aq}) \rightarrow \mathrm{Na}_2\mathrm{SO}_4(\mathrm{aq}) + 2\mathrm{H}_2\mathrm{O}(\mathrm{l})$$
Check: Na=2, O=6, H=4, S=1 — equal on both sides ✓

(c) $\mathrm{NaCl} + \mathrm{AgNO}_3 \rightarrow \mathrm{AgCl} + \mathrm{NaNO}_3$

• All atoms are already balanced: Na=1, Cl=1, Ag=1, N=1, O=3 on each side.

Balanced:
$$\mathrm{NaCl}(\mathrm{aq}) + \mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s})\downarrow + \mathrm{NaNO}_3(\mathrm{aq})$$
(Already balanced as written.)

(d) $\mathrm{BaCl}_2 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4 + \mathrm{HCl}$

• Cl: 2 on left → 2 HCl on right.
• H: 2 on left (H₂SO₄) → 2 HCl on right ✓

Balanced:
$$\mathrm{BaCl}_2(\mathrm{aq}) + \mathrm{H}_2\mathrm{SO}_4(\mathrm{aq}) \rightarrow \mathrm{BaSO}_4(\mathrm{s})\downarrow + 2\mathrm{HCl}(\mathrm{aq})$$
Check: Ba=1, Cl=2, H=2, S=1, O=4 — equal on both sides ✓

(a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water

Unbalanced: $\mathrm{Ca(OH)}_2 + \mathrm{CO}_2 \rightarrow \mathrm{CaCO}_3 + \mathrm{H}_2\mathrm{O}$

Check: Ca=1, C=1, O=4 left (2+2), O=4 right (3+1), H=2 on each side ✓

Balanced:
$$\mathrm{Ca(OH)}_2(\mathrm{aq}) + \mathrm{CO}_2(\mathrm{g}) \rightarrow \mathrm{CaCO}_3(\mathrm{s})\downarrow + \mathrm{H}_2\mathrm{O}(\mathrm{l})$$

(b) Zinc + Silver nitrate → Zinc nitrate + Silver

Unbalanced: $\mathrm{Zn} + \mathrm{AgNO}_3 \rightarrow \mathrm{Zn(NO}_3)_2 + \mathrm{Ag}$

• Zn(NO₃)₂ needs 2 NO₃ → 2 AgNO₃ on left → 2 Ag on right.

Balanced:
$$\mathrm{Zn}(\mathrm{s}) + 2\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{Zn(NO}_3)_2(\mathrm{aq}) + 2\mathrm{Ag}(\mathrm{s})$$
Check: Zn=1, Ag=2, N=2, O=6 — equal on both sides ✓

(c) Aluminium + Copper chloride → Aluminium chloride + Copper

Unbalanced: $\mathrm{Al} + \mathrm{CuCl}_2 \rightarrow \mathrm{AlCl}_3 + \mathrm{Cu}$

• AlCl₃ needs 3 Cl; CuCl₂ provides 2 Cl each → LCM of 3 and 2 = 6
• 2 AlCl₃ needs 6 Cl → 3 CuCl₂ on left → 3 Cu on right → 2 Al on left.

Balanced:
$$2\mathrm{Al}(\mathrm{s}) + 3\mathrm{CuCl}_2(\mathrm{aq}) \rightarrow 2\mathrm{AlCl}_3(\mathrm{aq}) + 3\mathrm{Cu}(\mathrm{s})$$
Check: Al=2, Cu=3, Cl=6 — equal on both sides ✓

(d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride

Unbalanced: $\mathrm{BaCl}_2 + \mathrm{K}_2\mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4 + \mathrm{KCl}$

• K: 2 on left → 2 KCl on right → Cl: 2 on right, 2 on left ✓

Balanced:
$$\mathrm{BaCl}_2(\mathrm{aq}) + \mathrm{K}_2\mathrm{SO}_4(\mathrm{aq}) \rightarrow \mathrm{BaSO}_4(\mathrm{s})\downarrow + 2\mathrm{KCl}(\mathrm{aq})$$
Check: Ba=1, Cl=2, K=2, S=1, O=4 — equal on both sides ✓

(a) Potassium bromide + Barium iodide → Potassium iodide + Barium bromide

Unbalanced: $\mathrm{KBr} + \mathrm{BaI}_2 \rightarrow \mathrm{KI} + \mathrm{BaBr}_2$

• Ba: 1 on each side ✓; BaBr₂ needs 2 Br → 2 KBr on left; 2 KI on right.

Balanced:
$$2\mathrm{KBr}(\mathrm{aq}) + \mathrm{BaI}_2(\mathrm{aq}) \rightarrow 2\mathrm{KI}(\mathrm{aq}) + \mathrm{BaBr}_2(\mathrm{s})$$
Check: K=2, Br=2, Ba=1, I=2 — equal on both sides ✓

Type of Reaction: Double Displacement Reaction (two compounds exchange their ions; BaBr₂ is precipitated).

(b) Zinc carbonate → Zinc oxide + Carbon dioxide

$$\mathrm{ZnCO}_3(\mathrm{s}) \xrightarrow{\Delta} \mathrm{ZnO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})$$

Check: Zn=1, C=1, O=3 on each side ✓ (Already balanced)

Type of Reaction: Decomposition Reaction (a single compound breaks down into two simpler substances on heating).

(c) Hydrogen + Chlorine → Hydrogen chloride

$$\mathrm{H}_2(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \rightarrow 2\mathrm{HCl}(\mathrm{g})$$

Check: H=2, Cl=2 on each side ✓

Type of Reaction: Combination Reaction (two elements combine to form a single product).

(d) Magnesium + Hydrochloric acid → Magnesium chloride + Hydrogen

Unbalanced: $\mathrm{Mg} + \mathrm{HCl} \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2$

• MgCl₂ needs 2 Cl → 2 HCl on left; H: 2 on left, 2 on right ✓

Balanced:
$$\mathrm{Mg}(\mathrm{s}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_2(\mathrm{aq}) + \mathrm{H}_2(\mathrm{g})\uparrow$$
Check: Mg=1, H=2, Cl=2 — equal on both sides ✓

Type of Reaction: Displacement Reaction (magnesium, being more reactive, displaces hydrogen from hydrochloric acid).