Chemical Reactions and Equations

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

✅ Answer (CBSE Exam-Ready):

Magnesium ribbon is cleaned with sandpaper before burning to remove the protective layer of magnesium oxide (MgO) that forms on its surface due to reaction with oxygen in air.

🧪 Why?

• Magnesium is very reactive.

• When exposed to air, it quickly forms a thin layer of MgO on its surface.

• This prevents direct contact of magnesium with oxygen during burning.

• Cleaning exposes the fresh metal, so it can burn easily and brightly in air.
  
  Certainly, Mayank! Below are the balanced chemical equations as per the Class 10 CBSE format — with and without state symbols where required. These are written in a clean, exam-ready style for classroom use or student worksheets.

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• ✍️ Q2: Write the balanced equations for the following chemical reactions.

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• (i) Hydrogen + Chlorine → Hydrogen chloride

• H2+Cl2→2HCl\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}

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

• 3BaCl2+Al2(SO4)3→3BaSO4+2AlCl33\text{BaCl}_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow 3\text{BaSO}_4 + 2\text{AlCl}_3

• (iii) Sodium + Water → Sodium hydroxide + Hydrogen

• 2Na+2H2O→2NaOH+H2↑2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 ↑

• ✍️ Q3: Write balanced chemical equations with state symbols.

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• (i) Barium chloride and sodium sulphate (both aqueous) react to form insoluble barium sulphate and sodium chloride solution

• BaCl2(aq)+Na2SO4(aq)→BaSO4(s)↓+2NaCl(aq)\text{BaCl}_2 (aq) + \text{Na}_2\text{SO}_4 (aq) \rightarrow \text{BaSO}_4 (s) ↓ + 2\text{NaCl} (aq)

• ✅ Type of reaction: Double displacement, precipitation
  ✅ BaSO₄ forms a white precipitate (insoluble in water)

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• (ii) Sodium hydroxide solution reacts with hydrochloric acid solution to form sodium chloride solution and water

• NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)\text{NaOH} (aq) + \text{HCl} (aq) \rightarrow \text{NaCl} (aq) + \text{H}_2\text{O} (l)

• ✅ Type of reaction: Neutralization reaction

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• Would you like me to turn these into a worksheet, MCQ quiz, or revision flashcards for students?

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Here are the exam-ready answers for your CBSE Class 10 Science questions based on Chapter 1: Chemical Reactions and Equations.

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

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

✅ The substance ‘X’ is quick lime.
Name: Calcium oxide
Formula: CaO

(ii) Write the reaction of substance ‘X’ with water.

When calcium oxide reacts with water, it forms calcium hydroxide, which is used for whitewashing.

CaO (s)+H2O (l)→Ca(OH)2(aq)+Heat\text{CaO (s)} + \text{H}_2\text{O (l)} \rightarrow \text{Ca(OH)}_2 (aq) + \text{Heat}

✅ This is an exothermic reaction (releases heat).

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

✅ In Activity 1.7, water is electrolysed to produce two gases:

• Hydrogen (H₂) at the cathode
  
  

• Oxygen (O₂) at the anode
  
  

✅ Answer:
The amount of gas collected in one test tube is double because water (H₂O) breaks down in a 2:1 ratio — 2 parts hydrogen and 1 part oxygen.

Chemical Equation:

2H2O (l)→electricity2H2(g)+O2(g)2\text{H}_2\text{O (l)} \xrightarrow{\text{electricity}} 2\text{H}_2 (g) + \text{O}_2 (g)

✅ Name of the gas collected in double amount:
Hydrogen gas (H₂)

Would you like these questions added to a chapter-end worksheet, formative assessment quiz, or animated class explanation for better concept delivery?

Certainly, Mayank. Here's a well-structured, CBSE-exam–oriented answer for all three questions, suitable for Class 10 Science students:

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

✅ Answer:

When an iron nail is dipped in copper sulphate (CuSO₄) solution, a displacement reaction occurs. Iron is more reactive than copper and displaces copper from its solution:

Fe(s)+CuSO4(aq)→FeSO4(aq)+Cu(s)\text{Fe} (s) + \text{CuSO}_4 (aq) \rightarrow \text{FeSO}_4 (aq) + \text{Cu} (s)

• The blue colour of the copper sulphate solution fades as iron sulphate (FeSO₄) is formed (which is pale green in colour).
  
  

• A reddish-brown deposit of copper forms on the iron nail.
  
  

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

✅ Answer:

Reaction between lead(II) nitrate and potassium iodide:

Pb(NO3)2(aq)+2KI(aq)→PbI2(s)+2KNO3(aq)\text{Pb(NO}_3)_2 (aq) + 2\text{KI} (aq) \rightarrow \text{PbI}_2 (s) + 2\text{KNO}_3 (aq)

• Lead iodide (PbI₂) is a yellow precipitate.
  
  

• This is a double displacement and precipitation reaction because the ions exchange to form a new insoluble product.
  
  

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

(i) 4Na(s)+O2(g)→2Na2O(s)4\text{Na}(s) + \text{O}_2(g) \rightarrow 2\text{Na}_2\text{O}(s)

• Sodium (Na) is oxidised (gains oxygen to form Na₂O).
  
  

• Oxygen (O₂) is reduced (is combined with sodium).
  
  

✅ Oxidised: Sodium (Na)
✅ Reduced: Oxygen (O₂)

(ii) CuO(s)+H2(g)→Cu(s)+H2O(l)\text{CuO}(s) + \text{H}_2(g) \rightarrow \text{Cu}(s) + \text{H}_2\text{O}(l)

• Copper oxide (CuO) is reduced (loses oxygen to become copper).
  
  

• Hydrogen (H₂) is oxidised (gains oxygen to form water).
  
  

✅ Oxidised: Hydrogen (H₂)
✅ Reduced: Copper oxide (CuO)

✅ Tip for Students:

To identify:

• Oxidation: Look for gain of oxygen / loss of hydrogen
  
  

• Reduction: Look for loss of oxygen / gain of hydrogen