1. Chemical Reactions and Equations Class 10 Science Solutions English Medium-NCERT Book Solutions

Questions: Page 6 (chap-1)

Q1. Why should a magnesium ribbon be cleaned before burning in air?

Ans: A magnesium ribbon should be cleaned before burning in air so that it may come in the contact of air.

Q2. Write the balanced equation for the following chemical reactions.

(i) Hydrogen + Chlorine → Hydrogen chloride

Sol:  H₂ + Cl₂ → HCl

      H₂ + Cl₂ → 2HCl

(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

Sol:  BaCl₂ + Al₂ (SO₄)₃ → Ba SO₄ + AlCl₃

      3 BaCl₂ + Al (SO₄)₃ → 3Ba SO₄ +2 AlCl₃

(iii) Sodium + Water → Sodium hydroxide + Hydrogen

Sol:  2Na + H₂O →  NaOH + H₂

      2Na + 2H₂O →  2NaOH + H₂

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.

Ans:  BaCl₂(aq) + Na₂ SO₄ (aq) → Ba SO₄ (s) + NaCl (aq)

(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

Ans:  NaOH(aq)  + HCl (aq) → NaCl (aq) + H₂O(aq)

Questions: Page 10 (chap-1)

Q1. A solution of a substance ‘X’ is used for white washing.

(i) Name the substance ‘X’ and write its formula.

Ans: Name the substance ‘X’ is Calsium Oxide (lime water) and its chemical formula is CaO.         

(ii) Write the reaction of the substance ‘X’ named in (i) above with water.

Ans: CaO + H₂O → Ca(OH)₂ 

Q2. Why is the amount of gas collected in one of the test tubes in Activity 1.7 double of the amount collected in the other? Name this gas.

Ans: During the Electrolysis of water, hydrogen and oxygen is get separated by the electricity. Water (H2O) contains two parts hydrogen and one part oxygen. Since hydrogen goes to one test tube and oxygen goes to another, the amount of gas collected in one of the test tubes is double of the amount collected in the other.

Questions: Page 13 (chap-1)

Q1. Why does the colour of copper sulphate solution change when an iron nail is dipped in it?

Ans: When an iron nail is dipped in a copper sulphate solution, iron which is more reactive than copper it displaces copper from copper sulphate solution and forms iron sulphate, In this case the blue colour of copper sulphate turns into green, which is colour of iron sulphate. 

Equation of this reaction:

 CuSO₄ + Fe → FeSO₄ + Cu 

(copper sulphate) (iron)  (iron sulphate) (copper)

Q2. Give an example of a double displacement reaction other than the one given in Activity 1.10.

Ans: Sodium carbonate reacts with calcium chloride to form calcium carbonate and sodium chloride.

Na₂CO₃ + CaCl₂ → CaCO₃ + 2 NaCl

In this reaction, sodium carbonate and calcium chloride exchange ions to form two new compounds. Hence, it is a double displacement reaction.

Q3.  Identify the substances that are oxidised and the substances that are reduced in the following reactions.

(i) 4Na(s) + O₂(g) → 2Na₂O(s)

(ii) CuO(s) + H₂(g) → Cu(s) + H₂O(l)

Ans: 

(i) Sodium (Na) is oxidised as it gains oxygen and oxygen gets reduced.
(ii).Copper oxide (CuO) is reduced to copper (Cu) while hydrogen (H2) gets oxidised to water (H2O).