Std 10th Science Part 1 – Chapter 3 : Chemical Reactions and Equations Exercise Questions And Answers Maharashtra Board

Q1. Choose the correct option from the bracket and explain the statement giving reason.

(Oxidation, displacement, electrolysis, reduction, zinc, copper, double displacement, decomposition)

a. To prevent rusting, a layer of …….. metal is applied on iron sheets.

Answer – To prevent rusting, a layer of zinc metal is applied on iron sheets.Reason: This process is called Galvanizing. Zinc is coated on iron because zinc is more reactive (electropositive) than iron. Zinc gets corroded (oxidized) first, protecting the iron underneath.

b. The conversion of ferrous sulphate to ferric sulphate is …….. reaction.

Answer – The conversion of ferrous sulphate to ferric sulphate is oxidation reaction.

2FeSO₄ → Fe₂(SO₄)₃

Fe₂ + SO₄^2- → 2Fe³⁺ + SO₄^2-

• Reason: In this conversion (Fe²⁺⟶Fe³⁺), the positive charge increases. This happens because the ferrous ion (Fe2+) loses an electron to form the ferric ion (Fe3+). Losing electrons is defined as oxidation.

c. When electric current is passed through acidulated water …….. of water takes place.

Answer – When electric current is passed through acidulated water electrolysis of water takes place.Reason: Electrolysis is a decomposition reaction that occurs due to electrical energy. Water decomposes into hydrogen gas and oxygen gas.

d. Addition of an aqueous solution of ZnSO4 to an aqueous solution of BaCl2 form a white precipitate is an example of ……. reaction.

Answer – Addition of an aqueous solution of ZnSO4​ to an aqueous solution of BaCl2​ to form a white precipitate is an example of double displacement reaction.

• Reason: In this reaction, the ions in the reactants are exchanged (Zn²⁺ exchanges with Ba²⁺ and SO₄²⁻​ exchanges with Cl⁻) to form a precipitate (insoluble solid, BaSO₄​). This is the definition of a double displacement reaction.

Q2. Write answers to the following.

a. What is the reaction called when oxidation and reduction take place simultaneously? Explain with one example.

Answer – Name: This reaction is called a Redox Reaction (Reduction + Oxidation).

Explanation: In a redox reaction, the reactant that gets oxidized (loses electrons/gains oxygen) is called the reductant, and the reactant that gets reduced (gains electrons/loses oxygen) is called the oxidant. They occur together.

Example:

• CuO is reduced (loses oxygen) to Cu.
• H_2​ is oxidized (gains oxygen) to H₂​O.

b. How can the rate of the chemical reaction, namely, decomposition of hydrogen peroxide be increased?

Answer – The rate of decomposition of hydrogen peroxide (2H₂​O₂​⟶2H₂​O+O₂​) can be increased by using a catalyst.

The catalyst used here is manganese dioxide (MnO_2​) powder. A catalyst is a substance that increases the rate of a chemical reaction without undergoing any chemical change itself.

c. Explain the term reactant and product giving examples.

Answer – 

Reactants: The substances that take part in a chemical reaction are called reactants. They are written on the left side of the chemical equation.

Products: The new substances that are formed as a result of a chemical reaction by forming new bonds are called products. They are written on the right side of the chemical equation.Example: In the reaction of combustion of coal:
C(s)+O₂​(g)⟶CO_2​(g)

• Reactants: Carbon (C) and Oxygen (O2​).
• Product: Carbon dioxide (CO2​).

d. Explain the types of reaction with reference to oxygen and hydrogen. Illustrate with examples.

Answer – Oxidation Reaction: A chemical reaction in which a reactant combines with oxygen or loses hydrogen to form the product.

• Example (Combining with oxygen):
  C+O₂​⟶CO_2​

Reduction Reaction: A chemical reaction in which a reactant gains hydrogen or loses oxygen to form the product.

• Example (Losing oxygen):
  CuO+H₂​⟶Cu+H₂​O

e. Explain the similarity and difference in two events, namely adding NaOH to water and adding CaO to water.

Answer – 

Similarity (Type of Process): Both are exothermic processes/reactions. In both cases, heat is given out (released), causing the temperature of the resulting mixture to increase.

CaO(s)+H₂​O(l)⟶Ca(OH)₂​(aq)+Heat

• Difference (Type of Change/Reaction):
  • NaOH to water: This is generally a physical change/process (dissolution). The sodium hydroxide solid simply dissolves in water.
  • CaO to water: This is a chemical reaction (combination reaction). Calcium oxide (slaked lime) reacts with water to form a completely new substance, calcium hydroxide, and heat is produced.

Q3. Explain the following terms with examples.

a. Endothermic reaction

Answer – Definition: A chemical reaction in which heat is absorbed from the surroundings or must be supplied continuously from outside.

Example: Decomposition of limestone.
CaCO₃​(s)+heat ⟶ CaO(s)+CO₂​(g)

b. Combination reaction:

Answer –

• Definition: A reaction in which two or more reactants combine to form a single product.

Example: Burning of magnesium in air to form magnesium oxide.
2Mg+O₂​⟶2MgO

c. Balanced equation:

Answer –

• Definition: A chemical equation in which the number of atoms of each element in the reactants (left side) is equal to the number of atoms of those elements in the products (right side). This follows the Law of Conservation of Mass.
• Example:
  2NaOH+H₂​SO₄​⟶Na_2​SO₄​+2H₂​O

d. Displacement reaction:

Answer –

• Definition: A reaction in which a more reactive element displaces a less reactive element’s ion from its compound.
• Example: Zinc displaces copper from copper sulphate.
  CuSO₄​(aq)+Zn(s)⟶ZnSO_4​(aq)+Cu(s)

Q4. Give scientific reasons. 

a.When the gas formed on heating limestone is passed through freshly prepared lime water, the lime water turns milky.

Answer – Reason: The gas formed on heating limestone (CaCO3​) is Carbon Dioxide (CO2).

When CO₂​ gas is passed through freshly prepared lime water (Ca(OH)₂​), it reacts to form insoluble white Calcium Carbonate (CaCO₃​), which makes the solution appear milky.

b. It takes time for pieces of Shahabad tile to disappear in HCl, but its powder disappears rapidly.

Answer – 

• Reason: The rate of a reaction depends on the size of the reactant particles.
• The powder form has a much larger surface area exposed for the reaction compared to large pieces.
• Smaller the size of the reactant particles, the higher is the rate of the reaction. Thus, the powder reacts (disappears) rapidly.

In the reaction of dil. HCl with pieces of Shahabad tile, CO₂ effervescence is formed amid the tile disappears slowly. On the other hand, CO₂ effervescence forms at a faster rate with Shahabad tile powder and it disappears rapidly.

c. While preparing dilute sulphuric acid from concentrated sulphuric acid in the laboratory, the concentrated sulphuric acid is added slowly to water with constant stirring.

Answer – 

• Reason: The dissolution of concentrated sulphuric acid in water is a highly exothermic process (releases a very large amount of heat).
• If water is poured into the concentrated acid, the concentrated acid will absorb the heat very slowly, and the heat liberated instantaneously might cause the water to
  evaporate violently. This can cause the concentrated acid to splatter and lead to a serious accident.
• By adding acid slowly to water while stirring, the heat is dispersed and only a small amount of heat is liberated at a time, preventing an accident.

d. It is recommended to use air tight container for storing oil for long time.

Answer – 

• Reason: Oil or food fried in oil, when exposed to air for a long time, undergoes air oxidation and becomes rancid (rancidity).
• Rancid food develops a foul odor and its taste changes.
• Storing oil/food in an air-tight container slows down the process of oxidation , thereby preventing rancidity and preserving the food for a longer time.

Q5. Observe the following picture a write down the chemical reaction with explanation. 

Answer – 

The picture shows the process of corrosion (rusting) of iron.

• Explanation: Corrosion of iron is an electrochemical reaction that requires both water (moisture) and air (oxygen). Different regions on the iron surface act as an anode and a cathode.
• Anode Region (Oxidation): Iron metal is oxidized to form ferrous ions (Fe²⁺).

Fe(s)⟶Fe²⁺(aq)+2e⁻

• Cathode Region (Reduction): Oxygen is reduced to form water (or hydroxide ions).

O₂​(g)+4H⁺(aq)+4e⁻⟶2H₂​O(l)

• Rust Formation: The Fe²⁺ ions then migrate and are further oxidized to Fe³⁺ ions. These
  Fe³⁺ ions react with water to form Reddish Hydrated Ferric Oxide (Fe₂​O₃​⋅xH₂​O), which is called rust.

2Fe³⁺(aq)+4H₂​O(l)⟶Fe₂​O₃​⋅H_2​O(s)+6H⁺(aq)

Q6. Identify from the following reactions the reactants that undergo oxidation and reduction.

a. Fe + S ⟶ FeS 

b. 2Ag₂O  ⟶ 4 Ag + O₂ ↑

c. 2Mg + O₂ ⟶ 2MgO d. NiO + H₂ ⟶ Ni + H₂ O

Answer – 

Reaction	Reactant Oxidized (Loses e− / Gains O)	Reactant Reduced (Gains e− / Loses O)
a. Fe+S⟶FeS	Fe (Loses e− to form Fe2+)	S (Gains e− to form S2−)
b. 2Ag2​O⟶4Ag+O2​↑	O2− in Ag2​O (Loses e− to form O2​)	Ag+ in Ag2​O (Loses O / Gains e−)
c. 2Mg+O2​⟶2MgO	Mg (Gains O)	O2​ (Gains e− to form O2−)
d. NiO+H2​⟶Ni+H2​O	H2​ (Gains O to form H2​O)	NiO (Loses O to form Ni)

Q7. Balance the following equation stepwise.

a. H₂​S₂​O₇​(l)+H₂​O(l)⟶H_2​SO_4​(l)

Answer – This reaction involves the combination of sulfur trioxide in sulfuric acid (H₂​S₂​O₇​, or Oleum) with water to form sulfuric acid.

Step 1: Count the atoms of each element on both sides.

Element	Reactants (H2​S2​O7​,H2​O)	Products (H2​SO4​)
H	2+2=4	2
S	2	1
O	7+1=8	4

Step 2: Balance Sulfur (S).

• Multiply H₂​SO₄​ (product) by 2 to get 2 S atoms.
  H₂​S₂​O₇​(l)+H₂​O(l)⟶2H₂​SO₄​(l)

Step 3: Final Check (All elements are balanced).

Element	Reactants (H2​S2​O7​,H2​O)	Products (H2​SO4​)
H	2+2=4	2×2=4
S	2	2×1=2
O	7+1=8	2×4=8

Balanced Equation:

H₂​S₂​O₇​(l)+H₂​O(l) ⟶ 2H₂​SO₄​(l)

b. SO_2​(g)+H₂​S(aq)⟶S(s)+H₂​O(l)

Answer – This is a redox reaction where sulfur dioxide reacts with hydrogen sulfide.

Step 1: Count the atoms of each element on both sides.

Element	Reactants (SO2​,H2​S)	Products (S,H2​O)
S	1+1=2	1
O	2	1
H	2	2

Step 2: Balance Oxygen (O).

• Multiply H₂​O (product) by 2 to get 2 O atoms.
  SO_2​(g)+H₂​S(aq)⟶S(s)+2H_2​O(l)

Step 3: Balance Hydrogen (H).

• The products now have 2×2=4 H atoms. The reactants have 2 H atoms (in H₂​S).
• Multiply H₂​S (reactant) by 2.
  SO_2​(g)+2H₂​S(aq)⟶S(s)+2H₂​O(l)

Step 4: Balance Sulfur (S) and Final Check.

• Reactants now have 1(in SO_2​)+2(in H₂​S)=3 S atoms.

Multiply S (product) by 3.
SO₂​(g)+2H₂​S(aq)⟶3S(s)+2H₂​O(l)

Element	Reactants (SO2​,H2​S)	Products (S,H2​O)
S	1+2=3	3
O	2	2×1=2
H	2×2=4	2×2=4

Balanced Equation:

SO₂​(g)+2H₂​S(aq)⟶3S(s)+2H₂​O(l)

c. Ag(s)+HCl(aq)⟶AgCl↓+H₂​↑

Answer – This reaction represents the attempt of a single metal (Ag) to displace hydrogen from an acid (HCl).

Step 1: Count the atoms of each element on both sides.

Element	Reactants (Ag,HCl)	Products (AgCl,H2​)
Ag	1	1
H	1	2
Cl	1	1

Step 2: Balance Hydrogen (H).

• Multiply HCl (reactant) by 2 to get 2 H atoms.
  Ag(s)+2HCl(aq)⟶AgCl↓+H₂​↑

Step 3: Balance Silver (Ag) and Chlorine (Cl).

• The reactants now have 2 Cl atoms, and the products have 1 Ag atom and 1 Cl atom (in AgCl).
• To balance Ag and Cl, multiply Ag (reactant) by 2 and AgCl (product) by 2.
  2Ag(s)+2HCl(aq)⟶2AgCl↓+H₂​↑

Step 4: Final Check.

Element	Reactants (Ag,HCl)	Products (AgCl,H2​)
Ag	2	2
H	2	2
Cl	2	2

Balanced Equation : 2Ag(s)+2HCl(aq)⟶2AgCl↓+H₂​↑

d. NaOH(aq)+H2​SO4​(aq)⟶Na2​SO4​(aq)+H2​O(l)

Answer – This is a neutralization reaction between a base (NaOH) and an acid (H₂​SO₄​).

Step 1: Count the atoms of each element on both sides.

Element	Reactants (NaOH,H2​SO4​)	Products (Na2​SO4​,H2​O)
Na	1	2
S	1	1
O	1+4=5	4+1=54+1=5
H	1+2=3	2

(Oxygen (O) and Sulfur (S) are already balanced.)

Step 2: Balance Sodium (Na).

• Multiply NaOH (reactant) by 2 to get 2 Na atoms.
  2NaOH(aq)+H₂​SO₄​(aq)⟶Na_2​SO₄​(aq)+H2​O(l)

Step 3: Balance Hydrogen (H) and Final Check.

• Count H: Reactants now have 2(in NaOH)+2(in H₂​SO₄​)=4 H atoms. Products have 2 H atoms (in H₂​O).
• Multiply H2​O (product) by 2 to get 4 H atoms.
  2NaOH(aq)+H₂​SO₄​(aq)⟶Na₂​SO₄​(aq)+2H₂​O(l)

Element	Reactants (NaOH,H2​SO4​)	Products (Na2​SO4​,H2​O)
Na	2	2
S	1	1
O	2(1)+4=6	4+2(1)=6
H	2(1)+2=4	2(2)=4

Balanced Equation:

2NaOH(aq)+H₂​SO₄​(aq)⟶Na₂​SO₄​(aq)+2H₂​O(l)

Q8. Identify the endothermic and exothermic reaction.

a. HCl+NaOH⟶NaCl+H₂​O+heat

• Answer: Exothermic (Heat is released/given out).

b. 2KClO₃​(s)⟶2KCl(s)+3O₂​↑

• Answer: Endothermic (Heat must be supplied for decomposition, as shown by Δ on the arrow in related decomposition reactions).

c. CaO+H₂​O⟶Ca(OH)₂​+heat

• Answer: Exothermic (Heat is released/given out).

d. CaCO₃​(s)⟶CaO(s)+CO₂​↑

• Answer: Endothermic (Heat must be supplied (Δ) for decomposition).

Q9. Match the column in the following table.

Reactants	Products	Type of chemical reaction
BaCl2​(aq)+ZnSO4​(aq)	H2​CO3​(aq)	Decomposition
2AgCl(s)	FeSO4​(aq)+Cu(s)	Double displacement
CuSO4​(aq)+Fe(s)	BaSO4​↓+ZnCl2​(aq)	Decomposition
H2​O(l)+CO2​(g)	2Ag(s)+Cl2​(g)	Double displacementCombination

Answer –

Reactants	Products	Type of chemical reaction
BaCl2​(aq)+ZnSO4​(aq)	BaSO4​↓+ZnCl2​(aq)	Double displacement
2AgCl(s)	2Ag(s)+Cl2​(g)	Decomposition
CuSO4​(aq)+Fe(s)	FeSO4​(aq)+Cu(s)	Decomposition
H2​O(l)+CO2​(g)	H2​CO3​(aq)	Combination