In-text Questions — Chemical Equations — Chemical Reactions and Equations

1. 11 markNCERT Cl-10 Science, In-text Qs after §1.1, Q1

   Why should a magnesium ribbon be cleaned before burning in air?

   Hint. Think about what happens to any reactive shiny metal surface if it's just left lying around in ordinary air for a while.

   Step 1 — What sits on the ribbon before it's even lit. Magnesium is a reactive metal, and a strip left exposed to air slowly reacts with atmospheric oxygen (and moisture), forming a thin layer of magnesium oxide on its surface.

   Step 2 — Why that layer is a problem. This oxide coating is chemically unreactive and will not burn. It sits between the flame and the actual metal underneath, since it is only the fresh magnesium metal that reacts vigorously with oxygen to give the dazzling white flame described in Activity 1.1.

   Step 3 — What cleaning does. Rubbing the ribbon with sandpaper scrapes off this oxide layer, exposing the shiny metal beneath, so it burns completely and cleanly: 2Mg(s) + O₂(g) → 2MgO(s).

   ✦ Answer: The ribbon develops a layer of magnesium oxide from slow reaction with atmospheric oxygen, and this layer will not burn. Cleaning it with sandpaper removes the oxide coating so the pure metal underneath is exposed and burns completely in air.

   Where students slip. Saying 'to remove dust or dirt' — the real reason is chemical (an oxide layer that has already formed), not a housekeeping one.

2. 23 marksNCERT Cl-10 Science, In-text Qs after §1.1, Q2

   Write the balanced equation for the following chemical reactions. (i) Hydrogen + Chlorine → Hydrogen chloride (ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride (iii) Sodium + Water → Sodium hydroxide + Hydrogen

   Hint. For (ii), balance whichever ion appears in the biggest 'chunk' first — here that's the sulphate group, since aluminium sulphate carries three of them at once.

   Step 1 — (i) Hydrogen + Chlorine → Hydrogen chloride. Skeletal: H₂ + Cl₂ → HCl. The left has 2 H and 2 Cl; the right has only 1 H and 1 Cl, so doubling the product fixes both at once: H₂(g) + Cl₂(g) → 2HCl(g).

   Step 2 — (ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride. Skeletal: BaCl₂ + Al₂(SO₄)₃ → BaSO₄ + AlCl₃. Aluminium sulphate supplies 2 Al and 3 sulphate groups, so 3 BaSO₄ are needed to use up all 3 sulphates — which in turn needs 3 BaCl₂ (giving 6 Cl). Those 6 Cl must reappear as 2 AlCl₃ (2 Al × 3 Cl = 6 Cl), which also matches the 2 Al already fixed: 3BaCl₂ + Al₂(SO₄)₃ → 3BaSO₄ + 2AlCl₃.

   Step 3 — (iii) Sodium + Water → Sodium hydroxide + Hydrogen. Skeletal: Na + H₂O → NaOH + H₂. Balancing hydrogen first: 2H₂O gives 4 H, matched by 2NaOH (2 H) + H₂ (2 H) = 4 H, so 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g), which also keeps Na and O equal.

   ✦ Answer: (i) H₂ + Cl₂ → 2HCl (ii) 3BaCl₂ + Al₂(SO₄)₃ → 3BaSO₄ + 2AlCl₃ (iii) 2Na + 2H₂O → 2NaOH + H₂

   Where students slip. In (ii), stopping after balancing barium and chlorine and forgetting to recheck aluminium and sulphate — a multi-ion equation like this one needs every element counted at the very end, not just the first one you balanced.

3. 32 marksNCERT Cl-10 Science, In-text Qs after §1.1, Q3

   Write a balanced chemical equation with state symbols for the following reactions. (i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride. (ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

   Hint. The word 'insoluble' in the question is telling you a state symbol directly.

   Step 1 — (i) Translate the words into a skeletal equation. Barium chloride (aq) + Sodium sulphate (aq) → Barium sulphate (insoluble ⇒ solid) + Sodium chloride (aq): BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + NaCl(aq).

   Step 2 — Balance it. Ba and S already match, but the right side has only 1 Na and 1 Cl against 2 of each on the left, so NaCl needs a coefficient of 2: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq).

   Step 3 — (ii) Translate and balance. NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l). Every element (Na, O, H, Cl) already appears once on each side, so no extra coefficients are needed here.

   ✦ Answer: (i) BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq) (ii) NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

   Where students slip. Marking the precipitate BaSO₄ as (aq) instead of (s) — the question already tells you it's insoluble, which means solid.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Science textbook, Reprint 2026-27 (jesc101.pdf) — three in-text question sets (8 questions total) plus one end-of-chapter Exercise (20 questions); the electron-transfer ('OIL RIG') definition of oxidation/reduction and the terms oxidising/reducing agent do not appear anywhere in the current chapter text. Questions are referenced from the NCERT textbook for identification.