Maharashtra State Board Solutions Class 10 Science Part 1 Ch 3

 Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 3 Chemical Reactions and Equations Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 3 Chemical Reactions and Equations

Question 1: Choose the correct option from the bracket and explain the statement giving reasons :
(Oxidation, displacement, electrolysis, reduction, zinc, copper, double diplacement, decomposition)

a. To prevent rusting, a laver of ……… metal is applied on iron sheets.
Answer: zinc

b. The conversion or ferrous sulphate to ferric sulphate is …….. reaction.
Answer: oxidation 

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

Answer: electrolysis

d. Addition of an aqueous solution of ZnSO4 to an aqueous solution of BaCl2 is an example of ……… reaction.
Answer: double displacement

Question 2:
a. What is the reaction called when oxidation and reduction take place simultaneously? Explain with one example.
Answer:
The reaction which involves simultaneous oxidation and reduction is called an oxidation-reduction or redox reaction.
In a redox reaction, one reactant gets oxidised while the other gets reduced during a reaction.
Redox reaction = Reduction + Oxidation

In redox reaction, the reductant is oxidized by the oxidant and the oxidant is reduced by the reductant.

Example:CuO(s) + H2(g) → Cu(s) + H2O
In this reaction, oxygen is removed from copper oxide therefore it is a reduction of CuO, while hydrogen accepts oxygen to form water that means oxidation of hydrogen takes place. Thus oxidation and reduction reactions occur simultaneously.

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

Answer:
At room temperature, the decomposition of hydrogen peroxide into water and oxygen takes place slowly. However, the same reaction occurs at a faster rate on adding manganese dioxide (MnO2),
powder in it.

c. Explain the term reactant product giving examples.
Answer:

1. The substance which undergoes bond breaking while taking part in a chemical reaction is called reactant.
2. The substance formed as a result of a chemical reaction by formation of new bonds is called product.
3. Example: For example: When two sodium atoms react with two chlorine atoms(reactants), they give a completely new compound (product) i.e. sodium chloride (two atoms).
   2Na+Cl2→2NaCl

d. Explain the types of reactions with reference to oxygen and hydrogen. Illustrate with examples.
Answer:
There are main three types of chemical reactions with reference to oxygen and hydrogen:
1.Combination Reaction
When two atoms react to form a compound, it is know as combination reaction.
For example: 2H2 + O2→ 2H2O$2H_2 + O_2\to 2H_{2}O$

2. Decomposition Reaction
When a compound breaks into simple molecular substances from which it is made up of, it is know as decomposition reaction.
For example: 2H2O→ 2H2 + O2$2H_{2}O\to 2H_2 + O_2$

3. Oxidation and reduction reaction:
Oxidation:
(i) The addition of oxygen to a substance is called oxidation.
(ii) The removal of hydrogen to a substance is called oxidation.
Reduction:
(i) The addition of hydrogen to a substance is called reduction.
(ii) The removal of oxygen to a substance is called reduction.
For example: CuO +H2→Cu + H2O$CuO +H_2\to Cu + H_{2}O$
In the above reaction, copper oxide is changing to Cu. That is, oxygen is being removed from copper oxide. So, copper oxide is being reduced to copper.
In the above reaction, H2 is changing into H2O. That is , oxygen is being added to hydrogen. So, hydrogen is being oxidised to water.

e. Explain the similarity and difference in two events, namely adding NaOH to water and adding CaO to water.
Answer:
Similarity : Both NaOH and CaO, when dissolved separately in water, solid NaOH dissolves releasing heat, resulting in rise in temperature. This reaction is exothermic reaction. When solid CaO dissolves in water, Ca(OH)2 is formed, large amount of heat is evolved. This reaction is also exothermic reaction. Both reactions are combination reactions and single product is obtained.
NaOH(s) + H2O → NaOH(aq) + Heat
CaO(s) + H2O → Ca(OH)2(aq) + Heat

Difference:

1. Aqueous solution of NaOH is considered as a strong alkali.
2. Aqueous solution of Ca(OH)2 is considered as a weak alkali.

Question 3: Explain the following terms with examples.

a. Endothermic reaction
Answer:
Endothermic reaction: The reaction in which heat is absorbed is called an endothermic
reaction.
when KNO3(s) dissolves in water, there is absorption of heat during the reaction and the temperature of the solution falls.
KNO2(s) + H2O(l) + Heat → KNO3(aq)

b. Combination reaction
Answer:
When two or more reactants combine in a reaction to form a single product, it is called a combination reaction.
For example: Magnesium and oxygen combine, when heated, to form magnesium oxide.
2Mg + O2 → 2MgO$2Mg + O_{2 }\to 2MgO$

c. Balanced equation
Answer:
In a chemical reaction, the number of atoms of the elements in the reactants is same as the number or atoms of those elements in the product, such an equation is called a balanced equation.
Example: AgNO3 + NaCl → AgCl + NaNO3
In the above reaction, the number of atoms of the elements in the reactants is same as the number of atoms of elements in the products.

d. Displacement reaction
Answer:
The reaction in which the place of the ion of a less reactive element in a compound is taken by another more reactive element by the formation of its own ions is called displacement reaction.

Example: When zinc granules are added to the blue coloured copper sulphate solution, the zinc ions formed from zinc atoms take the place of Cu²⁺ ions in CuSO₄ and copper atoms, formed from Cu²⁺ ions comes out i.e. the more reactive zinc displaces the less reactive Cu from copper sulphate.

Zn_(s) + CuSO_{4(aq) →} ZnSO_4(aq) + Cu_(s) + Heat

Question 4: Give scientific reason:

a. When the gas formed on heating lime stone is passed through freshly prepared lime water, the lime water turns milky.
Answer: when lime stone is heated, calcium oxide and carbon dioxide are formed. This carbon dioxide gas is passed through freshly prepared lime water, insoluble calcium carbonate and water are formed. In this reaction, lime water turns milky.
CaCo3→ CaO +CO2CO2+ Ca(OH)2→CaCO3

b. It takes time for pieces of Shahabud tile to disappear in HCl, but its powder disappears rapidly.
Answer:
The rate of a reaction depends upon the size of the particles of the reactants taking part in the reaction. The smaller the size of the reactants particles, the more is their total surface area and the faster is the rate of reaction.

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 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:
(1) The preparation of dilute sulphuric acid falls in the category of extreme exothermic process.

(2) During the preparation of dilute sulphuric acid. large amount of water is taken in a glass container which is surrounded by ice. Cool it for twenty minutes, Now small quantity of conc. H2SO4 is added slowly with stirring. Therefore, only a small amount of heat is liberated at a time. In this way dilute sulphuric acid is prepared.

(3) On the other hand, in the process of dilution or conc. sulphuric acid with water, very large amount of heat is liberated. As a result, water gets evaported instantaneously, if it is poured in to conc. H2SO4 which may cause an accident.

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

1. If edible oil is allowed to stand for a long time, it undergoes air oxidation, it becomes rancid and its smell and taste changes.
2. Rancidity in the rood stuff cooked in oil or ghee is prevented by using antioxidants. The process of oxidation reaction of food stuff can also be slowed down by storing it in air tight container.

Question 5: Observe the following picture a write down the chemical reaction with explanation.


Answer:
The rusting of iron is an oxidation process. The rust on iron does not form by a simple reaction between oxygen and iron surface. The rust is formed by an electrochemical reaction. Fe oxidises to Fe2O3. H2O on one part of iron surface while oxygen gets reduced to H2O on another part or surface, Different regions on the surface of iron become anode and cathode.
(1) Fe is oxidised to Fe2+ in the anode region.
Fe(s) → Fe2+ (aq) + 2e–
(2) O2 is reduced to form water in the cathode region.
O2(g) + 4H+ (aq) + 4e— → 2H2O(l)

When Fe2+ ions migrate from the anode region they react with water and futher get oxidised to form Fe3+ ions.
A reddish coloured hydrated oxide is formed from Fe3+ ions. It is called rust. It collects on the
surface.
2Fe3+ (aq) + 4H2O(l) → Fe2O3. H2O(s) + 6H+ (aq)…
Because of various components in the atmosphere, oxidation of metals takes place, consequently resulting in their damage. This is called ‘corrosion’. Iron rusts and a reddish coloured layer is formed on it. This is corrosion of iron.

Question 6: Identify from the following reactions the reactants that undergo oxidation and reduction.
a. Fe + S → FeS
Answer:
Fe + S → FeS
In this reaction, Iron (Fe) undergoes oxidation
and sulphur. (S) undergoes reduction.

b. 2Ag2O → 4Ag + O2↑
Answer:
2Ag2O → 4Ag + O2↑
In this reaction, reduction of Ag2O takes place.

c. 2Mg + O2 → 2MgO
Answer:
2Mg + O2 → 2MgO
In this reaction, oxidation of Mg takes place.

d. NiO + H2 → Ni + H2O
Answer:
NiO + H2 → Ni + H2O
In this reaction, reduction of NiO takes place and oxidation of H2 takes place.

Question 7: Balance the following equation stepwise.

a. H2S2O7(l) + H2O(l) → H2SO4(l)
Answer:
Step 1: Rewrite the given equation as it is
H2S2O7(l) + H2O(l) → H2SO4(l)
Step 2: write the number or atoms of each element in the unbalanced equation on both sides of equations.

Element	Number of atoms in reactant (left side)	Number of atoms in products (right side)
H	4	2
S	2	1
O	8	4

Step 3: To equalise the number of hydrogen atoms, sulphur atoms and oxygen atoms we use 2 as the coemficient or factor in the product.

Element	Number of atoms in reactant (left side)	Number of atoms in products (right side)
H	4	2 × 2
S	2	1 × 2
O	8	4 × 2
Total	14	14

Now the equation becomes H2S2O7 + H2O → 2H2SO4
Now, count the atoms of each element on both sides of the equation. The number of atoms on both sides are equal. Hence, the balanced equation is
H2S2O7 + H2O → 2H2SO4
Now indicate the physical states of the reactants and products.
H2S2O7(l) + H2O(l) → 2H2SO4(l)

b. SO2(g) + H2S(aq) → S(s) + H2O(l)
Answer:
Step 1:
Rewrite the given equation as it is
SO2(g) + H2S(aq) → S(s) + H2O(l)

Step 2:
Write the number of atoms of each element in the unbalanced equation on both sides of equations.

Element	Number of atoms in reactants (left side)	Number of atoms in products (right side)
S	2	1
O	2	1
H	2	2

The number of hydrogen atoms on both sides of the equation is same, therefore, equalise the number of sulphur atoms and oxygen atoms.

Step 3: To balance the number of sulphur atoms:

Number of atoms of sulphur	In reactants		In products
	S2O	H2S	(S)
Initially	1	1	1
To balance	1	1	1 × 2

To equalise the number of sulphur atoms, we use 2 as the factor in the product, now the equation becomes
SO2 + H2S → 2S + H2O

Step 4:
To equalise the number of oxygen atoms in the unbalanced equation.

Number of atoms of oxygen	In reactants (SO2)	In products H2O
Initially	2	1
To balance	2	1 × 2

To equalise the number of sulphur atoms, we use 2 as the factor in the product i.e. H2O, now the unbalanced equation becomes
SO2 + H2S → 2S + 2H2O

Step 5:
To equalise the number of hydrogen atoms in unbalanced equation:

Number of atoms of hydrogen	In reactants (H2S)	In products (H2O)
Initially	2	4
To balance	2 × 2	4

To equalise the number of hydrogen atoms we use 2 as the factor in the reactant i.e, H2S, now the unbalanced equation become
SO2 + 2H2S → 2S + 2H2O
Now, count the atoms of each element on both sides of the equation, there are less number of sulphur atoms in the product. Now equalise the sulphur atoms, the balanced equation becomes,
SO2 + 2H2S → 3S + 2H2O
Now indicate the physical states of reactants and products.
SO2(g) + 2H2S(aq) → 3S(s) + 2H2O(l)

c. Ag(s) + HCl(l) → AgCl ↓ + H2 ↑
Answer:
Step 1:
Rewrite the given equation as it is
Ag(s) + HCl(l) → AgCl ↓ + H2 ↑

Step 2:
write the number of atoms or each element in the unbalanced equation on both sides of equations.

Element	Number of atoms in reactants (left side)	Number of atoms in products (right side)
Ag	1	1
H	1	2
Cl	1	1

The number of silver and chlorine atoms on both sides of the equation are same, therefore, equalise the number of hydrogen atoms.

Step 3:
To balance the number of hydrogen atoms.

Number of atoms of hydrogen	In reactants HCl	In products H2
Initially	1	2
To balance	1 × 2	2

To equalise the number of hydrogen atoms, we use 2 as the factor in the product HCl, now the unbalanced equation become
Ag(s) + 2HCl → AgCl + H2

Step 4:
To balance the number of chlorine atoms:

Number of atoms of chlorine	In reactants (2HCl)	In products (AgCl)
Initially	2	1
To balance	2	2 ×1

To equalise the number of chlorine atoms, we use 2 as the factor in the product AgCl. now the unbalanced equation becomes
Ag + 2HCl → 2AgCl + H2
Now count the atoms of each element on both sides of the equation, there are less number of silver atoms in the reactant. Now equalise the silver atoms, the balanced equation becomes
2Ag + 2HCl → 2AgCl + H2
Now indicate the physical states of the reactunts and products
2Ag(s) + 2HCl(l) → 2AgCl ↓ + H2 ↑

d. H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l)
Answer:
Step 1:
Rewrite the given equation as it is
H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l)

Step 2:
write the number of atoms of each element in the unbalanced equation on both sides of the equation.

Element	Number of atoms in reactants	Number of atoms in products
Na	1	2
S	1	1
O	5	5
H	3	2

The number of oxygen atoms involved in different compounds on both sides (reactants and products) are equal. Therefore, balance the number of atoms of the second element, sodium.

Step 3:
To balance the number of sodium atoms:

Number of atoms of sodium	In reactants	In products
To begin with	1 (in NaOH)	2 (in Na2SO4)
To balance	1 × 2	2

To equalise the number of sodium atoms, we use 2 as the factor of NaOH in the reactants. Now, the partly balanced equation becomes as follows
H2SO4 + 2NaOH → Na2SO4 + H2O

Step 4:
Now, balance the number of hydrogen atoms:

Number of atoms of hydrogen	In reactants	In products
To begin with	(in H2SO4) 2 (in NaOH)	2 (in H2O)
To balance	4	2 × 2

To equalise the number of hydrogen atoms, we use 2 as the factor or H2O in the products. The equation then becomes
H2SO4 + 2NaOH → Na2SO4 + H2O
Now, count the atoms of each element on both sides of the equation. The number of atoms on both sides are equal. Hence, the balanced equation is
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Now indicate the physical states of the reactants and the products.
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

Question 8: Identify the endothermic and exothermic reaction.
a. HCl + NaOH → NaCl + H2O + heat
Answer:
Exothermic reaction.

b. 2KClO3(s)⟶Δ2KCl(s)+3O2↑2KClO3(s)⟶Δ2KCl(s)+3O2↑
Answer: Exothermic reaction.

c. CaO + H2O → Ca(OH)2 + heat
Answer: Exothermic reaction.

d. CaCO3(s)⟶ΔCaO(s)+CO2↑CaCO3(s)⟶ΔCaO(s)+CO2↑
Answer: Exothermic reaction.

Question 9: Match the column in the following table:

Reactants	products	Type of chemical reaction
BaCl2(aq) + ZnSO4(aq)	H2CO3(aq)	Displacement
2 AgCl(s)	FeSO4(aq) + Cu(s)	Combination
CuSO4(aq) + Fe(s)	BaSO4↓ + ZnCl2(aq)	Decomposition
H2O(l) + CO2(g)	2Ag(s) + Cl2(g)	Double displacement

Answer:

Reactants	products	Type of chemical reaction
BaCl2(aq) + ZnSO4(aq)	BaSO4↓ + ZnCl2(aq)	Double displacement
2 AgCl(s)	2Ag(s) + Cl2(g)	Decomposition
CuSO4(aq) + Fe(s)	FeSO4(aq) + Cu(s)	Displacement
H2O(l) + CO2(g)	H2CO3(aq)	Combination

Project:
Do it your self:
1. Prepare aqueous solutions or various solid salts available in the laboratory. Observe what happens when aqueous solution of sodium hydroxide is added to these. Prepare a chart of double displacement reactions based on these observation.

2. Observe and note the physical and chemical changes experienced in various incidents in your day to day 1ife.